Does normality of a subgroup imply it has index 2?
I know that if $G$ is a group, $N < G$, then the condition that $|G:N|=2$ implies that$ N$ is normal in $G$. But what about the converse if we know that $N$ is normal in $G$ does that then imply that the index of $N$ in $G$ is $2$ ?
For some context from the author of this MSE question, see this comment:
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in Sp and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G – can'tcauchy
group-theory normal-subgroups
edited Dec 27 '18 at 15:22
Shaun
8,711113680
asked Dec 26 '18 at 18:26
can'tcauchy
973417
|
show 2 more comments
I know that if $G$ is a group, $N < G$, then the condition that $|G:N|=2$ implies that$ N$ is normal in $G$. But what about the converse if we know that $N$ is normal in $G$ does that then imply that the index of $N$ in $G$ is $2$ ?
For some context from the author of this MSE question, see this comment:
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in Sp and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G – can'tcauchy
group-theory normal-subgroups
edited Dec 27 '18 at 15:22
Shaun
8,711113680
asked Dec 26 '18 at 18:26
can'tcauchy
973417
2
Hint 1: every subgroup of an abelian group is normal. Hint 2: the identity element is a normal subgroup of every group (abelian or not).
– lulu
Dec 26 '18 at 18:29
2
Hint. The subgroup containing just the identity is normal.
– Ethan Bolker
Dec 26 '18 at 18:29
1
Whoever voted to close, would you mind explaining how this question is off topic ?
– can'tcauchy
Dec 26 '18 at 18:49
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in $S_p$ and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G
– can'tcauchy
Dec 26 '18 at 19:54
Very good! So it is really worth to try for yourself. Perhaps now the posted question is somehow obsolete (sounds better than off topic).
– Dietrich Burde
Dec 26 '18 at 20:13
|
show 2 more comments
I know that if $G$ is a group, $N < G$, then the condition that $|G:N|=2$ implies that$ N$ is normal in $G$. But what about the converse if we know that $N$ is normal in $G$ does that then imply that the index of $N$ in $G$ is $2$ ?
For some context from the author of this MSE question, see this comment:
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in Sp and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G – can'tcauchy
group-theory normal-subgroups
edited Dec 27 '18 at 15:22
Shaun
8,711113680
asked Dec 26 '18 at 18:26
can'tcauchy
973417
I know that if $G$ is a group, $N < G$, then the condition that $|G:N|=2$ implies that$ N$ is normal in $G$. But what about the converse if we know that $N$ is normal in $G$ does that then imply that the index of $N$ in $G$ is $2$ ?
For some context from the author of this MSE question, see this comment:
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in Sp and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G – can'tcauchy
group-theory normal-subgroups
group-theory normal-subgroups
edited Dec 27 '18 at 15:22
Shaun
8,711113680
asked Dec 26 '18 at 18:26
can'tcauchy
973417
edited Dec 27 '18 at 15:22
Shaun
8,711113680
asked Dec 26 '18 at 18:26
can'tcauchy
973417
edited Dec 27 '18 at 15:22
Shaun
8,711113680
edited Dec 27 '18 at 15:22
Shaun
8,711113680
edited Dec 27 '18 at 15:22
Shaun
8,711113680
8,711113680
asked Dec 26 '18 at 18:26
can'tcauchy
973417
asked Dec 26 '18 at 18:26
can'tcauchy
973417
asked Dec 26 '18 at 18:26
can'tcauchy
973417
973417
2
Hint 1: every subgroup of an abelian group is normal. Hint 2: the identity element is a normal subgroup of every group (abelian or not).
– lulu
Dec 26 '18 at 18:29
2
Hint. The subgroup containing just the identity is normal.
– Ethan Bolker
Dec 26 '18 at 18:29
1
Whoever voted to close, would you mind explaining how this question is off topic ?
– can'tcauchy
Dec 26 '18 at 18:49
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in $S_p$ and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G
– can'tcauchy
Dec 26 '18 at 19:54
Very good! So it is really worth to try for yourself. Perhaps now the posted question is somehow obsolete (sounds better than off topic).
– Dietrich Burde
Dec 26 '18 at 20:13
|
show 2 more comments
2
Hint 1: every subgroup of an abelian group is normal. Hint 2: the identity element is a normal subgroup of every group (abelian or not).
– lulu
Dec 26 '18 at 18:29
2
Hint. The subgroup containing just the identity is normal.
– Ethan Bolker
Dec 26 '18 at 18:29
1
Whoever voted to close, would you mind explaining how this question is off topic ?
– can'tcauchy
Dec 26 '18 at 18:49
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in $S_p$ and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G
– can'tcauchy
Dec 26 '18 at 19:54
Very good! So it is really worth to try for yourself. Perhaps now the posted question is somehow obsolete (sounds better than off topic).
– Dietrich Burde
Dec 26 '18 at 20:13
2
2
Hint 1: every subgroup of an abelian group is normal. Hint 2: the identity element is a normal subgroup of every group (abelian or not).
– lulu
Dec 26 '18 at 18:29
Hint 1: every subgroup of an abelian group is normal. Hint 2: the identity element is a normal subgroup of every group (abelian or not).
– lulu
Dec 26 '18 at 18:29
2
2
Hint. The subgroup containing just the identity is normal.
– Ethan Bolker
Dec 26 '18 at 18:29
Hint. The subgroup containing just the identity is normal.
– Ethan Bolker
Dec 26 '18 at 18:29
1
1
Whoever voted to close, would you mind explaining how this question is off topic ?
– can'tcauchy
Dec 26 '18 at 18:49
Whoever voted to close, would you mind explaining how this question is off topic ?
– can'tcauchy
Dec 26 '18 at 18:49
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in $S_p$ and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G
– can'tcauchy
Dec 26 '18 at 19:54
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in $S_p$ and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G
– can'tcauchy
Dec 26 '18 at 19:54
Very good! So it is really worth to try for yourself. Perhaps now the posted question is somehow obsolete (sounds better than off topic).
– Dietrich Burde
Dec 26 '18 at 20:13
Very good! So it is really worth to try for yourself. Perhaps now the posted question is somehow obsolete (sounds better than off topic).
– Dietrich Burde
Dec 26 '18 at 20:13
|
show 2 more comments
1 Answer
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Every group is a normal subgroup of itself.
If you insist that the subgroup is not the group itself, then the subgroup containing only the identity is normal in every group.
If you insist that the subgroup is proper, then consider constructing a group by means of a semidirect product. In particular, as @egreg points out in the comments below, a direct product is easiest.
answered Dec 26 '18 at 18:30
Shaun
8,711113680
1
ah okay and then that has index 1, so the implication only goes one way . thank you :)
– can'tcauchy
Dec 26 '18 at 18:33
You're welcome, @can'tcauchy; please don't forget to accept this answer using the checkmark $checkmark$.
– Shaun
Dec 26 '18 at 18:34
2
@Shaun A direct product is much simpler: take a group $G$ of whatever cardinality and a group $H$ of cardinality $n$; then $Gtimes{1}$ is a normal subgroup of $Gtimes H$ of index $n$.
– egreg
Dec 26 '18 at 18:45
@Shaun oh I will , it just said I have to wait a minute or two :). By the way on your final comment about constructing the group from a semi direct product then one of the groups must be normal and the other just a subgroup but in that case ( unless I'm mistaken ) we have that half the group has elements from N right ? in that case the index will be two right ?
– can'tcauchy
Dec 26 '18 at 18:45
1
@Shaun never mind egreg answered my question
– can'tcauchy
Dec 26 '18 at 18:48
|
show 1 more comment
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Every group is a normal subgroup of itself.
If you insist that the subgroup is not the group itself, then the subgroup containing only the identity is normal in every group.
If you insist that the subgroup is proper, then consider constructing a group by means of a semidirect product. In particular, as @egreg points out in the comments below, a direct product is easiest.
answered Dec 26 '18 at 18:30
Shaun
8,711113680
1
ah okay and then that has index 1, so the implication only goes one way . thank you :)
– can'tcauchy
Dec 26 '18 at 18:33
You're welcome, @can'tcauchy; please don't forget to accept this answer using the checkmark $checkmark$.
– Shaun
Dec 26 '18 at 18:34
2
@Shaun A direct product is much simpler: take a group $G$ of whatever cardinality and a group $H$ of cardinality $n$; then $Gtimes{1}$ is a normal subgroup of $Gtimes H$ of index $n$.
– egreg
Dec 26 '18 at 18:45
@Shaun oh I will , it just said I have to wait a minute or two :). By the way on your final comment about constructing the group from a semi direct product then one of the groups must be normal and the other just a subgroup but in that case ( unless I'm mistaken ) we have that half the group has elements from N right ? in that case the index will be two right ?
– can'tcauchy
Dec 26 '18 at 18:45
1
@Shaun never mind egreg answered my question
– can'tcauchy
Dec 26 '18 at 18:48
|
show 1 more comment
Every group is a normal subgroup of itself.
If you insist that the subgroup is not the group itself, then the subgroup containing only the identity is normal in every group.
If you insist that the subgroup is proper, then consider constructing a group by means of a semidirect product. In particular, as @egreg points out in the comments below, a direct product is easiest.
answered Dec 26 '18 at 18:30
Shaun
8,711113680
1
ah okay and then that has index 1, so the implication only goes one way . thank you :)
– can'tcauchy
Dec 26 '18 at 18:33
You're welcome, @can'tcauchy; please don't forget to accept this answer using the checkmark $checkmark$.
– Shaun
Dec 26 '18 at 18:34
2
@Shaun A direct product is much simpler: take a group $G$ of whatever cardinality and a group $H$ of cardinality $n$; then $Gtimes{1}$ is a normal subgroup of $Gtimes H$ of index $n$.
– egreg
Dec 26 '18 at 18:45
@Shaun oh I will , it just said I have to wait a minute or two :). By the way on your final comment about constructing the group from a semi direct product then one of the groups must be normal and the other just a subgroup but in that case ( unless I'm mistaken ) we have that half the group has elements from N right ? in that case the index will be two right ?
– can'tcauchy
Dec 26 '18 at 18:45
1
@Shaun never mind egreg answered my question
– can'tcauchy
Dec 26 '18 at 18:48
|
show 1 more comment
Every group is a normal subgroup of itself.
If you insist that the subgroup is not the group itself, then the subgroup containing only the identity is normal in every group.
If you insist that the subgroup is proper, then consider constructing a group by means of a semidirect product. In particular, as @egreg points out in the comments below, a direct product is easiest.
answered Dec 26 '18 at 18:30
Shaun
8,711113680
Every group is a normal subgroup of itself.
If you insist that the subgroup is not the group itself, then the subgroup containing only the identity is normal in every group.
If you insist that the subgroup is proper, then consider constructing a group by means of a semidirect product. In particular, as @egreg points out in the comments below, a direct product is easiest.
answered Dec 26 '18 at 18:30
Shaun
8,711113680
edited Dec 27 '18 at 23:21
answered Dec 26 '18 at 18:30
Shaun
8,711113680
answered Dec 26 '18 at 18:30
Shaun
8,711113680
answered Dec 26 '18 at 18:30
Shaun
8,711113680
8,711113680
1
ah okay and then that has index 1, so the implication only goes one way . thank you :)
– can'tcauchy
Dec 26 '18 at 18:33
You're welcome, @can'tcauchy; please don't forget to accept this answer using the checkmark $checkmark$.
– Shaun
Dec 26 '18 at 18:34
2
@Shaun A direct product is much simpler: take a group $G$ of whatever cardinality and a group $H$ of cardinality $n$; then $Gtimes{1}$ is a normal subgroup of $Gtimes H$ of index $n$.
– egreg
Dec 26 '18 at 18:45
@Shaun oh I will , it just said I have to wait a minute or two :). By the way on your final comment about constructing the group from a semi direct product then one of the groups must be normal and the other just a subgroup but in that case ( unless I'm mistaken ) we have that half the group has elements from N right ? in that case the index will be two right ?
– can'tcauchy
Dec 26 '18 at 18:45
1
@Shaun never mind egreg answered my question
– can'tcauchy
Dec 26 '18 at 18:48
|
show 1 more comment
1
ah okay and then that has index 1, so the implication only goes one way . thank you :)
– can'tcauchy
Dec 26 '18 at 18:33
You're welcome, @can'tcauchy; please don't forget to accept this answer using the checkmark $checkmark$.
– Shaun
Dec 26 '18 at 18:34
2
@Shaun A direct product is much simpler: take a group $G$ of whatever cardinality and a group $H$ of cardinality $n$; then $Gtimes{1}$ is a normal subgroup of $Gtimes H$ of index $n$.
– egreg
Dec 26 '18 at 18:45
@Shaun oh I will , it just said I have to wait a minute or two :). By the way on your final comment about constructing the group from a semi direct product then one of the groups must be normal and the other just a subgroup but in that case ( unless I'm mistaken ) we have that half the group has elements from N right ? in that case the index will be two right ?
– can'tcauchy
Dec 26 '18 at 18:45
1
@Shaun never mind egreg answered my question
– can'tcauchy
Dec 26 '18 at 18:48
1
1
ah okay and then that has index 1, so the implication only goes one way . thank you :)
– can'tcauchy
Dec 26 '18 at 18:33
ah okay and then that has index 1, so the implication only goes one way . thank you :)
– can'tcauchy
Dec 26 '18 at 18:33
You're welcome, @can'tcauchy; please don't forget to accept this answer using the checkmark $checkmark$.
– Shaun
Dec 26 '18 at 18:34
You're welcome, @can'tcauchy; please don't forget to accept this answer using the checkmark $checkmark$.
– Shaun
Dec 26 '18 at 18:34
2
2
@Shaun A direct product is much simpler: take a group $G$ of whatever cardinality and a group $H$ of cardinality $n$; then $Gtimes{1}$ is a normal subgroup of $Gtimes H$ of index $n$.
– egreg
Dec 26 '18 at 18:45
@Shaun A direct product is much simpler: take a group $G$ of whatever cardinality and a group $H$ of cardinality $n$; then $Gtimes{1}$ is a normal subgroup of $Gtimes H$ of index $n$.
– egreg
Dec 26 '18 at 18:45
@Shaun oh I will , it just said I have to wait a minute or two :). By the way on your final comment about constructing the group from a semi direct product then one of the groups must be normal and the other just a subgroup but in that case ( unless I'm mistaken ) we have that half the group has elements from N right ? in that case the index will be two right ?
– can'tcauchy
Dec 26 '18 at 18:45
@Shaun oh I will , it just said I have to wait a minute or two :). By the way on your final comment about constructing the group from a semi direct product then one of the groups must be normal and the other just a subgroup but in that case ( unless I'm mistaken ) we have that half the group has elements from N right ? in that case the index will be two right ?
– can'tcauchy
Dec 26 '18 at 18:45
1
1
@Shaun never mind egreg answered my question
– can'tcauchy
Dec 26 '18 at 18:48
@Shaun never mind egreg answered my question
– can'tcauchy
Dec 26 '18 at 18:48
|
show 1 more comment
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Hint 1: every subgroup of an abelian group is normal. Hint 2: the identity element is a normal subgroup of every group (abelian or not).
– lulu
Dec 26 '18 at 18:29
2
Hint. The subgroup containing just the identity is normal.
– Ethan Bolker
Dec 26 '18 at 18:29
1
Whoever voted to close, would you mind explaining how this question is off topic ?
– can'tcauchy
Dec 26 '18 at 18:49
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in $S_p$ and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G
– can'tcauchy
Dec 26 '18 at 19:54
Very good! So it is really worth to try for yourself. Perhaps now the posted question is somehow obsolete (sounds better than off topic).
– Dietrich Burde
Dec 26 '18 at 20:13