Intersection of a Family of Normal Subgroups of a Group G is also a Normal Subgroup of G
$begingroup$
I have this classic exercise to prove:
If $ { N_i : i in I } $ is a family of normal subgroups of a group $ G $, then $ bigcap_{i in I} N_i $ is a normal subgroup of $ G $ too.
I am searching for a constructive reasoning.
My solution is like as follows:
Let's explicite the set of indices: $ I_n = { 1,2,...,n } $. And let's give a name to the set mentioned above: $ F_i = { N_i : forall i in I_n, N_i < G } $. The crucial difference is that: we will suppose that elements of our set $ F $ are not normal subgroups of G. Because we know that this enunciation is a valid property when $ N_i $ are ordinary subgroups.
Then define a function $ f : F_i to wp (G) $ such that, for $ i = 1 $ we have $ f(1) = N_1 $, and for $ i ge 2 $, we have $ f(i) = N_i bigcap f(N_{i-1}) $.
We can easily see that: $ forall i in I_n, f(i) subset wp (G) $, even better: $ f(i) < G $. As we can check: $ Dom(f) = F_i $ is a group on its own under the usual law of intersection; and the same for $ Im(f) = wp (G) $.
So the tactic is to check if this function $ f $ is a homomorphism of groups? If it is so, then $ f $ preserves algebraic properties, and, as we started with the known conclusion (the case where sets are not normal subgroups) and constructed our correspondence, we can conclude that the weaker property will be inherited.
My need for help begins here: I don't know if $ f $ is well defined, AND, I couldn't prove that my function $ f $ is indeed a homomorphism.
Thank you all in advance for any advice and correction.
abstract-algebra group-theory normal-subgroups
asked Dec 22 '18 at 0:50
freehumoristfreehumorist
351214
$endgroup$
|
show 6 more comments
$begingroup$
I have this classic exercise to prove:
If $ { N_i : i in I } $ is a family of normal subgroups of a group $ G $, then $ bigcap_{i in I} N_i $ is a normal subgroup of $ G $ too.
I am searching for a constructive reasoning.
My solution is like as follows:
Let's explicite the set of indices: $ I_n = { 1,2,...,n } $. And let's give a name to the set mentioned above: $ F_i = { N_i : forall i in I_n, N_i < G } $. The crucial difference is that: we will suppose that elements of our set $ F $ are not normal subgroups of G. Because we know that this enunciation is a valid property when $ N_i $ are ordinary subgroups.
Then define a function $ f : F_i to wp (G) $ such that, for $ i = 1 $ we have $ f(1) = N_1 $, and for $ i ge 2 $, we have $ f(i) = N_i bigcap f(N_{i-1}) $.
We can easily see that: $ forall i in I_n, f(i) subset wp (G) $, even better: $ f(i) < G $. As we can check: $ Dom(f) = F_i $ is a group on its own under the usual law of intersection; and the same for $ Im(f) = wp (G) $.
So the tactic is to check if this function $ f $ is a homomorphism of groups? If it is so, then $ f $ preserves algebraic properties, and, as we started with the known conclusion (the case where sets are not normal subgroups) and constructed our correspondence, we can conclude that the weaker property will be inherited.
My need for help begins here: I don't know if $ f $ is well defined, AND, I couldn't prove that my function $ f $ is indeed a homomorphism.
Thank you all in advance for any advice and correction.
abstract-algebra group-theory normal-subgroups
asked Dec 22 '18 at 0:50
freehumoristfreehumorist
351214
$endgroup$
$begingroup$
What is $wp (G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:54
$begingroup$
It is the Power Set of G. I couldn't find the appropriate symbol on Jax.
$endgroup$
– freehumorist
Dec 22 '18 at 0:55
$begingroup$
And what is the assumed group operation on $wp(G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:56
3
$begingroup$
You're overcomplicating this. You also make a mistake in the very first sentence: the indices needn't be countable. If each $N_i$ is normal, it follows that for each $g in G$ and each $n_i in N_i$, we have $gn_ig^{-1} in N_i$. Hence if $n^{*} in bigcap N_i$ and $g in G$, what can we say about $gn^{*}g^{-1}$?
$endgroup$
– MathematicsStudent1122
Dec 22 '18 at 0:57
$begingroup$
Usual intersection. We need not Subsets of G be Subgroups. We need closure, associativity, etc. for Power Set of G to be a Group only.
$endgroup$
– freehumorist
Dec 22 '18 at 0:58
|
show 6 more comments
$begingroup$
I have this classic exercise to prove:
If $ { N_i : i in I } $ is a family of normal subgroups of a group $ G $, then $ bigcap_{i in I} N_i $ is a normal subgroup of $ G $ too.
I am searching for a constructive reasoning.
My solution is like as follows:
Let's explicite the set of indices: $ I_n = { 1,2,...,n } $. And let's give a name to the set mentioned above: $ F_i = { N_i : forall i in I_n, N_i < G } $. The crucial difference is that: we will suppose that elements of our set $ F $ are not normal subgroups of G. Because we know that this enunciation is a valid property when $ N_i $ are ordinary subgroups.
Then define a function $ f : F_i to wp (G) $ such that, for $ i = 1 $ we have $ f(1) = N_1 $, and for $ i ge 2 $, we have $ f(i) = N_i bigcap f(N_{i-1}) $.
We can easily see that: $ forall i in I_n, f(i) subset wp (G) $, even better: $ f(i) < G $. As we can check: $ Dom(f) = F_i $ is a group on its own under the usual law of intersection; and the same for $ Im(f) = wp (G) $.
So the tactic is to check if this function $ f $ is a homomorphism of groups? If it is so, then $ f $ preserves algebraic properties, and, as we started with the known conclusion (the case where sets are not normal subgroups) and constructed our correspondence, we can conclude that the weaker property will be inherited.
My need for help begins here: I don't know if $ f $ is well defined, AND, I couldn't prove that my function $ f $ is indeed a homomorphism.
Thank you all in advance for any advice and correction.
abstract-algebra group-theory normal-subgroups
asked Dec 22 '18 at 0:50
freehumoristfreehumorist
351214
$endgroup$
I have this classic exercise to prove:
If $ { N_i : i in I } $ is a family of normal subgroups of a group $ G $, then $ bigcap_{i in I} N_i $ is a normal subgroup of $ G $ too.
I am searching for a constructive reasoning.
My solution is like as follows:
Let's explicite the set of indices: $ I_n = { 1,2,...,n } $. And let's give a name to the set mentioned above: $ F_i = { N_i : forall i in I_n, N_i < G } $. The crucial difference is that: we will suppose that elements of our set $ F $ are not normal subgroups of G. Because we know that this enunciation is a valid property when $ N_i $ are ordinary subgroups.
Then define a function $ f : F_i to wp (G) $ such that, for $ i = 1 $ we have $ f(1) = N_1 $, and for $ i ge 2 $, we have $ f(i) = N_i bigcap f(N_{i-1}) $.
We can easily see that: $ forall i in I_n, f(i) subset wp (G) $, even better: $ f(i) < G $. As we can check: $ Dom(f) = F_i $ is a group on its own under the usual law of intersection; and the same for $ Im(f) = wp (G) $.
So the tactic is to check if this function $ f $ is a homomorphism of groups? If it is so, then $ f $ preserves algebraic properties, and, as we started with the known conclusion (the case where sets are not normal subgroups) and constructed our correspondence, we can conclude that the weaker property will be inherited.
My need for help begins here: I don't know if $ f $ is well defined, AND, I couldn't prove that my function $ f $ is indeed a homomorphism.
Thank you all in advance for any advice and correction.
abstract-algebra group-theory normal-subgroups
abstract-algebra group-theory normal-subgroups
asked Dec 22 '18 at 0:50
freehumoristfreehumorist
351214
asked Dec 22 '18 at 0:50
freehumoristfreehumorist
351214
asked Dec 22 '18 at 0:50
freehumoristfreehumorist
351214
asked Dec 22 '18 at 0:50
freehumoristfreehumorist
351214
asked Dec 22 '18 at 0:50
freehumoristfreehumorist
351214
351214
$begingroup$
What is $wp (G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:54
$begingroup$
It is the Power Set of G. I couldn't find the appropriate symbol on Jax.
$endgroup$
– freehumorist
Dec 22 '18 at 0:55
$begingroup$
And what is the assumed group operation on $wp(G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:56
3
$begingroup$
You're overcomplicating this. You also make a mistake in the very first sentence: the indices needn't be countable. If each $N_i$ is normal, it follows that for each $g in G$ and each $n_i in N_i$, we have $gn_ig^{-1} in N_i$. Hence if $n^{*} in bigcap N_i$ and $g in G$, what can we say about $gn^{*}g^{-1}$?
$endgroup$
– MathematicsStudent1122
Dec 22 '18 at 0:57
$begingroup$
Usual intersection. We need not Subsets of G be Subgroups. We need closure, associativity, etc. for Power Set of G to be a Group only.
$endgroup$
– freehumorist
Dec 22 '18 at 0:58
|
show 6 more comments
$begingroup$
What is $wp (G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:54
$begingroup$
It is the Power Set of G. I couldn't find the appropriate symbol on Jax.
$endgroup$
– freehumorist
Dec 22 '18 at 0:55
$begingroup$
And what is the assumed group operation on $wp(G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:56
3
$begingroup$
You're overcomplicating this. You also make a mistake in the very first sentence: the indices needn't be countable. If each $N_i$ is normal, it follows that for each $g in G$ and each $n_i in N_i$, we have $gn_ig^{-1} in N_i$. Hence if $n^{*} in bigcap N_i$ and $g in G$, what can we say about $gn^{*}g^{-1}$?
$endgroup$
– MathematicsStudent1122
Dec 22 '18 at 0:57
$begingroup$
Usual intersection. We need not Subsets of G be Subgroups. We need closure, associativity, etc. for Power Set of G to be a Group only.
$endgroup$
– freehumorist
Dec 22 '18 at 0:58
$begingroup$
What is $wp (G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:54
$begingroup$
What is $wp (G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:54
$begingroup$
It is the Power Set of G. I couldn't find the appropriate symbol on Jax.
$endgroup$
– freehumorist
Dec 22 '18 at 0:55
$begingroup$
It is the Power Set of G. I couldn't find the appropriate symbol on Jax.
$endgroup$
– freehumorist
Dec 22 '18 at 0:55
$begingroup$
And what is the assumed group operation on $wp(G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:56
$begingroup$
And what is the assumed group operation on $wp(G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:56
3
3
$begingroup$
You're overcomplicating this. You also make a mistake in the very first sentence: the indices needn't be countable. If each $N_i$ is normal, it follows that for each $g in G$ and each $n_i in N_i$, we have $gn_ig^{-1} in N_i$. Hence if $n^{*} in bigcap N_i$ and $g in G$, what can we say about $gn^{*}g^{-1}$?
$endgroup$
– MathematicsStudent1122
Dec 22 '18 at 0:57
$begingroup$
You're overcomplicating this. You also make a mistake in the very first sentence: the indices needn't be countable. If each $N_i$ is normal, it follows that for each $g in G$ and each $n_i in N_i$, we have $gn_ig^{-1} in N_i$. Hence if $n^{*} in bigcap N_i$ and $g in G$, what can we say about $gn^{*}g^{-1}$?
$endgroup$
– MathematicsStudent1122
Dec 22 '18 at 0:57
$begingroup$
Usual intersection. We need not Subsets of G be Subgroups. We need closure, associativity, etc. for Power Set of G to be a Group only.
$endgroup$
– freehumorist
Dec 22 '18 at 0:58
$begingroup$
Usual intersection. We need not Subsets of G be Subgroups. We need closure, associativity, etc. for Power Set of G to be a Group only.
$endgroup$
– freehumorist
Dec 22 '18 at 0:58
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Much simpler: take $;xinbigcap_{iin I} N_i;$ , then for any $;gin G;$ :
$$forall,iin I;,;;x^g:=gxg^{-1}in N_iimplies x^ginbigcap _{iin I}N_i;ldots$$
Observe that in the above it isn't assumed the index set $;I;$ is countable: that doesn't affect.
answered Dec 22 '18 at 1:25
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Much simpler: take $;xinbigcap_{iin I} N_i;$ , then for any $;gin G;$ :
$$forall,iin I;,;;x^g:=gxg^{-1}in N_iimplies x^ginbigcap _{iin I}N_i;ldots$$
Observe that in the above it isn't assumed the index set $;I;$ is countable: that doesn't affect.
answered Dec 22 '18 at 1:25
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
$begingroup$
Much simpler: take $;xinbigcap_{iin I} N_i;$ , then for any $;gin G;$ :
$$forall,iin I;,;;x^g:=gxg^{-1}in N_iimplies x^ginbigcap _{iin I}N_i;ldots$$
Observe that in the above it isn't assumed the index set $;I;$ is countable: that doesn't affect.
answered Dec 22 '18 at 1:25
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
$begingroup$
Much simpler: take $;xinbigcap_{iin I} N_i;$ , then for any $;gin G;$ :
$$forall,iin I;,;;x^g:=gxg^{-1}in N_iimplies x^ginbigcap _{iin I}N_i;ldots$$
Observe that in the above it isn't assumed the index set $;I;$ is countable: that doesn't affect.
answered Dec 22 '18 at 1:25
DonAntonioDonAntonio
180k1494233
$endgroup$
Much simpler: take $;xinbigcap_{iin I} N_i;$ , then for any $;gin G;$ :
$$forall,iin I;,;;x^g:=gxg^{-1}in N_iimplies x^ginbigcap _{iin I}N_i;ldots$$
Observe that in the above it isn't assumed the index set $;I;$ is countable: that doesn't affect.
answered Dec 22 '18 at 1:25
DonAntonioDonAntonio
180k1494233
answered Dec 22 '18 at 1:25
DonAntonioDonAntonio
180k1494233
answered Dec 22 '18 at 1:25
DonAntonioDonAntonio
180k1494233
answered Dec 22 '18 at 1:25
DonAntonioDonAntonio
180k1494233
180k1494233
add a comment |
add a comment |
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$begingroup$
What is $wp (G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:54
$begingroup$
It is the Power Set of G. I couldn't find the appropriate symbol on Jax.
$endgroup$
– freehumorist
Dec 22 '18 at 0:55
$begingroup$
And what is the assumed group operation on $wp(G)$?
$endgroup$
– Bungo
Dec 22 '18 at 0:56
3
$begingroup$
You're overcomplicating this. You also make a mistake in the very first sentence: the indices needn't be countable. If each $N_i$ is normal, it follows that for each $g in G$ and each $n_i in N_i$, we have $gn_ig^{-1} in N_i$. Hence if $n^{*} in bigcap N_i$ and $g in G$, what can we say about $gn^{*}g^{-1}$?
$endgroup$
– MathematicsStudent1122
Dec 22 '18 at 0:57
$begingroup$
Usual intersection. We need not Subsets of G be Subgroups. We need closure, associativity, etc. for Power Set of G to be a Group only.
$endgroup$
– freehumorist
Dec 22 '18 at 0:58