Permutations and terminology












1












$begingroup$


Say I have the following permutation



$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$



which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.



What is the name for such transformation?



Thank you






permutations permutation-cycles







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:10












  • $begingroup$
    Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:19












  • $begingroup$
    Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:20








  • 1




    $begingroup$
    Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:23








  • 1




    $begingroup$
    You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 19:31
















1












$begingroup$


Say I have the following permutation



$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$



which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.



What is the name for such transformation?



Thank you






permutations permutation-cycles







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:10












  • $begingroup$
    Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:19












  • $begingroup$
    Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:20








  • 1




    $begingroup$
    Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:23








  • 1




    $begingroup$
    You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 19:31














1












1








1





$begingroup$


Say I have the following permutation



$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$



which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.



What is the name for such transformation?



Thank you






permutations permutation-cycles







share|cite|improve this question











$endgroup$




Say I have the following permutation



$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$



which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.



What is the name for such transformation?



Thank you






permutations permutation-cycles




permutations permutation-cycles






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 19:25









Bernard

119k740113




119k740113










asked Dec 2 '18 at 19:09









Adam54Adam54

515




515








  • 2




    $begingroup$
    The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:10












  • $begingroup$
    Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:19












  • $begingroup$
    Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:20








  • 1




    $begingroup$
    Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:23








  • 1




    $begingroup$
    You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 19:31














  • 2




    $begingroup$
    The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:10












  • $begingroup$
    Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:19












  • $begingroup$
    Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:20








  • 1




    $begingroup$
    Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:23








  • 1




    $begingroup$
    You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 19:31








2




2




$begingroup$
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:10






$begingroup$
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:10














$begingroup$
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
$endgroup$
– Adam54
Dec 2 '18 at 19:19






$begingroup$
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
$endgroup$
– Adam54
Dec 2 '18 at 19:19














$begingroup$
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:20






$begingroup$
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:20






1




1




$begingroup$
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:23






$begingroup$
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:23






1




1




$begingroup$
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 19:31




$begingroup$
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 19:31










1 Answer
1






active

oldest

votes


















0












$begingroup$

$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49
















0












$begingroup$

$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49














0












0








0





$begingroup$

$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.






share|cite|improve this answer











$endgroup$



$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 21:32

























answered Dec 2 '18 at 19:26









amWhyamWhy

1




1












  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49


















  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49
















$begingroup$
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
$endgroup$
– Adam54
Dec 2 '18 at 19:29




$begingroup$
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
$endgroup$
– Adam54
Dec 2 '18 at 19:29












$begingroup$
Top number goes to bottom number in each column.
$endgroup$
– amWhy
Dec 2 '18 at 19:37




$begingroup$
Top number goes to bottom number in each column.
$endgroup$
– amWhy
Dec 2 '18 at 19:37












$begingroup$
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
$endgroup$
– Adam54
Dec 2 '18 at 19:42




$begingroup$
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
$endgroup$
– Adam54
Dec 2 '18 at 19:42












$begingroup$
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
$endgroup$
– amWhy
Dec 2 '18 at 19:46




$begingroup$
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
$endgroup$
– amWhy
Dec 2 '18 at 19:46












$begingroup$
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
$endgroup$
– Adam54
Dec 2 '18 at 19:49




$begingroup$
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
$endgroup$
– Adam54
Dec 2 '18 at 19:49


















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Bundesstraße 106

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Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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Verónica Boquete

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Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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