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Show that the group generated by $(23)$ in $S_3$ is not normal.

My approach to this problem is the following:

Let $H=\langle (23) \rangle$. If $H$ was normal in $S_3$, then it would satisfy the property that $gH=Hg$ for all $g \in S_3$. The elements of $S_3$ are $\{(1),(12),(13),(23),(123),(132)\}$

I know that I need to show that this condition fails to solve the problem, but I am not sure what elements are in $H$.

How do I proceed in this?

Thank you for your time and assistance.

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  • $\begingroup$ Well, $(23)(23)=(1)$, if $H$ is generated by $(23)$, then $H=$? $\endgroup$ – Ben Sheller Oct 8 '15 at 19:15
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You have $$(1 \ 2) (2 \ 3) (1 \ 2 )^{-1} = (1 \ 2) (2 \ 3) (1 \ 2 )= (1 \ 3) \notin H$$ So $H$ is not normal.

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Conjugation in $S_n$ is about relabelling the objects that you are permuting. If $\sigma, \tau \in S_n$, then $\sigma^{-1}\tau\sigma$ is the permutation that works by (1) relabelling everything using $\sigma$, (2) permuting the relabelled objects with $\tau$ and then (3) undoing the relabelling using $\sigma^{-1}$. So in $S_n$ (for $n \ge 3$), the subgroup generated by a single transposition $(2\,3)$ won't be fixed under conjugation, because you can relabel so that the objects being permuted are not $2$ and $3$.

(The above assumes that you are writing functions on the left of their arguments: switch $\sigma$ and $\sigma^{-1}$ if you prefer to write functions to the right of their arguments.)

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As $(23)(23)=(1)$ the only elements in $H$ are $(1)$ and $(23)$.

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  • $\begingroup$ "I know that I need to show that this condition fails to solve the problem, but I am not sure what elements are in $H$.How do I proceed in this?" This is what I answered. Now that this stumbling block is removed and they know what $H$ is they could just do the check they said they need to do. $\endgroup$ – quid Oct 8 '15 at 21:29

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