Let $G$ be a group & $S=\{s_1,s_2,\dots,s_k\}$ generate $G$. Let $H\le G$. Then if $s_iH s_i^{−1}\subseteq H\forall s_i\in S,H$ is normal

Asked Feb 26 '21 at 22:00

Active Feb 26 '21 at 22:30

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Let $G$ be a group and let $S = \{s_1,s_2,\dots,s_k\}$ be a generating set of $G$. Assume $H$ is a subgroup. Then if $s_i H s_i^{−1}\subseteq H \forall s_i\in S,H$ is normal.

I was trying to solve a few questions from a practice sheet but I couldn't really understand how to do it.

I tried contraposition and contradiction and I couldn't really get anywhere. I also tried to show that $s_i^{-1}H s_i$ is a subset of $H$ but honestly got nowhere with that. Any help is appreciated.

edited Feb 26 '21 at 22:11

Shaun

44,997

asked Feb 26 '21 at 22:00

mathfanatic

Given any element $g\in G$, write it as a product of elements of $S$. And then expand $g=s_{i_1}s_{i_2}\ldots s_{i_n}$ in $gHg^{-1}$ and evaluate in to out. – Rushabh Mehta Feb 26 '21 at 22:05
You cannot show that $s_i^{-1}Hs$ is a subset of $H$, because there are examples in which it is not true. – Derek Holt Feb 26 '21 at 22:08
@DonThousand The problem with that is that you might need $s_i^{-1}$ as well as $s_i$ to express $g$. – Derek Holt Feb 26 '21 at 22:09
@DerekHolt True, can't edit my comment though. The procedure is still functionally the same. – Rushabh Mehta Feb 26 '21 at 22:10
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In the example given by user1729 in this post, the hypothesis holds with $S = {a, t^{-1}}$, but the subgroup $H$ is not normal. – Derek Holt Feb 26 '21 at 22:19
Thanks for your help, @DerekHolt. – mathfanatic Feb 26 '21 at 22:24
Here's another example that also essentially shows that this is not sufficient for normality. – Arturo Magidin Feb 27 '21 at 00:10

Let $G$ be a group & $S=\{s_1,s_2,\dots,s_k\}$ generate $G$. Let $H\le G$. Then if $s_iH s_i^{−1}\subseteq H\forall s_i\in S,H$ is normal

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