Show that a group of order 66 has a normal subgroup of order 33.
$begingroup$
This question is somewhat similar to: A group of order $66$ has an element of order $33$.
However, I do not understand how I would show that the subgroup of order 33 is normal. So far I have that there is a unique Sylow 11-subgroup, but I don't know how to continue.
Any hints are appreciated.
group-theory finite-groups sylow-theory
asked Dec 29 '18 at 23:43
AnonAnon
113
$endgroup$
add a comment |
$begingroup$
This question is somewhat similar to: A group of order $66$ has an element of order $33$.
However, I do not understand how I would show that the subgroup of order 33 is normal. So far I have that there is a unique Sylow 11-subgroup, but I don't know how to continue.
Any hints are appreciated.
group-theory finite-groups sylow-theory
asked Dec 29 '18 at 23:43
AnonAnon
113
$endgroup$
7
$begingroup$
Every subgroup of index $2$ is normal - see for example here.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:45
1
$begingroup$
Related; math.stackexchange.com/questions/84632/…
$endgroup$
– Servaes
Dec 29 '18 at 23:46
$begingroup$
I would say that your question is fully covered by this more general result. I won't vote to close right away, because there is scope for disagreement. I have the dupehammer on group-theory, so my vote would be immediately binding.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:17
add a comment |
$begingroup$
This question is somewhat similar to: A group of order $66$ has an element of order $33$.
However, I do not understand how I would show that the subgroup of order 33 is normal. So far I have that there is a unique Sylow 11-subgroup, but I don't know how to continue.
Any hints are appreciated.
group-theory finite-groups sylow-theory
asked Dec 29 '18 at 23:43
AnonAnon
113
$endgroup$
This question is somewhat similar to: A group of order $66$ has an element of order $33$.
However, I do not understand how I would show that the subgroup of order 33 is normal. So far I have that there is a unique Sylow 11-subgroup, but I don't know how to continue.
Any hints are appreciated.
group-theory finite-groups sylow-theory
group-theory finite-groups sylow-theory
asked Dec 29 '18 at 23:43
AnonAnon
113
asked Dec 29 '18 at 23:43
AnonAnon
113
edited Dec 29 '18 at 23:50
Anon
asked Dec 29 '18 at 23:43
AnonAnon
113
asked Dec 29 '18 at 23:43
AnonAnon
113
asked Dec 29 '18 at 23:43
AnonAnon
113
113
7
$begingroup$
Every subgroup of index $2$ is normal - see for example here.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:45
1
$begingroup$
Related; math.stackexchange.com/questions/84632/…
$endgroup$
– Servaes
Dec 29 '18 at 23:46
$begingroup$
I would say that your question is fully covered by this more general result. I won't vote to close right away, because there is scope for disagreement. I have the dupehammer on group-theory, so my vote would be immediately binding.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:17
add a comment |
7
$begingroup$
Every subgroup of index $2$ is normal - see for example here.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:45
1
$begingroup$
Related; math.stackexchange.com/questions/84632/…
$endgroup$
– Servaes
Dec 29 '18 at 23:46
$begingroup$
I would say that your question is fully covered by this more general result. I won't vote to close right away, because there is scope for disagreement. I have the dupehammer on group-theory, so my vote would be immediately binding.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:17
7
7
$begingroup$
Every subgroup of index $2$ is normal - see for example here.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:45
$begingroup$
Every subgroup of index $2$ is normal - see for example here.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:45
1
1
$begingroup$
Related; math.stackexchange.com/questions/84632/…
$endgroup$
– Servaes
Dec 29 '18 at 23:46
$begingroup$
Related; math.stackexchange.com/questions/84632/…
$endgroup$
– Servaes
Dec 29 '18 at 23:46
$begingroup$
I would say that your question is fully covered by this more general result. I won't vote to close right away, because there is scope for disagreement. I have the dupehammer on group-theory, so my vote would be immediately binding.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:17
$begingroup$
I would say that your question is fully covered by this more general result. I won't vote to close right away, because there is scope for disagreement. I have the dupehammer on group-theory, so my vote would be immediately binding.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $P$ be the Sylow $11$-subgroup you've shown exists. Since you've shown $P$ is unique, we have that $P$ is normal. Now if we let $Q$ be a Sylow $3$-subgroup, then $PQ=QP$ by normality of $P$, and thus $PQ$ is a subgroup of $G$. But $PQ$ has order $33$ (hence index $2$) so $PQ$ is normal in $G$.
answered Dec 29 '18 at 23:55
Alex MathersAlex Mathers
11.3k21444
$endgroup$
add a comment |
$begingroup$
We know by the linked answer that there exists an element of order $33$. Let this element be $x$. Hence, $langle xrangle$ has order $33$. We know that all subgroups with index 2 in a group are normal.
edited Dec 30 '18 at 0:08
MJD
47.9k29217398
answered Dec 29 '18 at 23:52
Don ThousandDon Thousand
4,574734
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $P$ be the Sylow $11$-subgroup you've shown exists. Since you've shown $P$ is unique, we have that $P$ is normal. Now if we let $Q$ be a Sylow $3$-subgroup, then $PQ=QP$ by normality of $P$, and thus $PQ$ is a subgroup of $G$. But $PQ$ has order $33$ (hence index $2$) so $PQ$ is normal in $G$.
answered Dec 29 '18 at 23:55
Alex MathersAlex Mathers
11.3k21444
$endgroup$
add a comment |
$begingroup$
Let $P$ be the Sylow $11$-subgroup you've shown exists. Since you've shown $P$ is unique, we have that $P$ is normal. Now if we let $Q$ be a Sylow $3$-subgroup, then $PQ=QP$ by normality of $P$, and thus $PQ$ is a subgroup of $G$. But $PQ$ has order $33$ (hence index $2$) so $PQ$ is normal in $G$.
answered Dec 29 '18 at 23:55
Alex MathersAlex Mathers
11.3k21444
$endgroup$
add a comment |
$begingroup$
Let $P$ be the Sylow $11$-subgroup you've shown exists. Since you've shown $P$ is unique, we have that $P$ is normal. Now if we let $Q$ be a Sylow $3$-subgroup, then $PQ=QP$ by normality of $P$, and thus $PQ$ is a subgroup of $G$. But $PQ$ has order $33$ (hence index $2$) so $PQ$ is normal in $G$.
answered Dec 29 '18 at 23:55
Alex MathersAlex Mathers
11.3k21444
$endgroup$
Let $P$ be the Sylow $11$-subgroup you've shown exists. Since you've shown $P$ is unique, we have that $P$ is normal. Now if we let $Q$ be a Sylow $3$-subgroup, then $PQ=QP$ by normality of $P$, and thus $PQ$ is a subgroup of $G$. But $PQ$ has order $33$ (hence index $2$) so $PQ$ is normal in $G$.
answered Dec 29 '18 at 23:55
Alex MathersAlex Mathers
11.3k21444
answered Dec 29 '18 at 23:55
Alex MathersAlex Mathers
11.3k21444
answered Dec 29 '18 at 23:55
Alex MathersAlex Mathers
11.3k21444
answered Dec 29 '18 at 23:55
Alex MathersAlex Mathers
11.3k21444
11.3k21444
add a comment |
add a comment |
$begingroup$
We know by the linked answer that there exists an element of order $33$. Let this element be $x$. Hence, $langle xrangle$ has order $33$. We know that all subgroups with index 2 in a group are normal.
edited Dec 30 '18 at 0:08
MJD
47.9k29217398
answered Dec 29 '18 at 23:52
Don ThousandDon Thousand
4,574734
$endgroup$
add a comment |
$begingroup$
We know by the linked answer that there exists an element of order $33$. Let this element be $x$. Hence, $langle xrangle$ has order $33$. We know that all subgroups with index 2 in a group are normal.
edited Dec 30 '18 at 0:08
MJD
47.9k29217398
answered Dec 29 '18 at 23:52
Don ThousandDon Thousand
4,574734
$endgroup$
add a comment |
$begingroup$
We know by the linked answer that there exists an element of order $33$. Let this element be $x$. Hence, $langle xrangle$ has order $33$. We know that all subgroups with index 2 in a group are normal.
edited Dec 30 '18 at 0:08
MJD
47.9k29217398
answered Dec 29 '18 at 23:52
Don ThousandDon Thousand
4,574734
$endgroup$
We know by the linked answer that there exists an element of order $33$. Let this element be $x$. Hence, $langle xrangle$ has order $33$. We know that all subgroups with index 2 in a group are normal.
edited Dec 30 '18 at 0:08
MJD
47.9k29217398
answered Dec 29 '18 at 23:52
Don ThousandDon Thousand
4,574734
edited Dec 30 '18 at 0:08
MJD
47.9k29217398
edited Dec 30 '18 at 0:08
MJD
47.9k29217398
edited Dec 30 '18 at 0:08
MJD
47.9k29217398
47.9k29217398
answered Dec 29 '18 at 23:52
Don ThousandDon Thousand
4,574734
answered Dec 29 '18 at 23:52
Don ThousandDon Thousand
4,574734
answered Dec 29 '18 at 23:52
Don ThousandDon Thousand
4,574734
4,574734
add a comment |
add a comment |
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7
$begingroup$
Every subgroup of index $2$ is normal - see for example here.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:45
1
$begingroup$
Related; math.stackexchange.com/questions/84632/…
$endgroup$
– Servaes
Dec 29 '18 at 23:46
$begingroup$
I would say that your question is fully covered by this more general result. I won't vote to close right away, because there is scope for disagreement. I have the dupehammer on group-theory, so my vote would be immediately binding.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 14:17