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Suppose that $H$ is a subgroup of $G$ and that $h$ and $h'$ are in $H$. If $h$ and $h'$ are conjugates in $G$, are they also conjugates in $H$? Why or why not? I have proved that the converse is true.

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Let's consider symmetric group of order $3,$ $$S_3=\{(1),(123)(132)(12)(23)(13)\}$$ and its alternating subgroup $$A_3=\{(1),(123),(132)\}.$$ Note that $(123)$ and $(132)$ are conjugate in $S_3$ but not in $A_3.$

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  • $\begingroup$ Yes, they are: $(132) = (123)(123)(123)^{-1}$. $\endgroup$ – rogerl Oct 25 '14 at 13:53
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    $\begingroup$ @rogerl:$(123)(123)(123)^{−1}=(123)$ $\endgroup$ – Bumblebee Oct 25 '14 at 13:56
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    $\begingroup$ Usually if I want a counter example, I'm looking at $S_3$ and most of the times it works. $\endgroup$ – Bumblebee Oct 25 '14 at 14:26
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No, is the simple answer.

For a small counterexample I would suggest looking for a situation where $H$ is abelian and $G$ is not. Then each element of $H$ is conjugate only to itself within $H$, but may have other conjugates when elements of $G$ are taken into account.

I would explore this a bit yourself to see if you can find an example which works.


Since another answer has suggested looking at the non-abelian group of order $6$ which we can identify with $S_3$, the symmetric group on three elements, there are some features of this which bear a little reflection.

In any symmetric group the conjugacy classes are determined by the cycle types of the permutations, so we can find examples quite easily in any symmetric group on more than two elements.

Here, if we take $A_3$ for $H$ we can note that this is a normal subgroup and is abelian. Its elements are the identity and both the elements of order $3$. The action of $S_3$ on $A_3$ by conjugation involves the outer automorphism of $A_3$ which exchanges the two elements of order $3$.

So examples can be found by taking a group $H$ and embedding it in a larger group which realises at least one outer automorphism. One way of doing this is by the semidirect product.

There are however groups which have no outer automorphisms - and this is true for $S_n$ except for $S_6$ (the outer automorphism of $S_6$ is itself an interesting thing to explore). So if $H$ were $S_5$, for example, we could say that every pair of elements of $H$ conjugate in $G$ are also conjugate in $H$.

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The answer must clearly be no. An easy way to construct counterexamples is to let $h_0,h_1\in H$ be such that some automorphism $\phi$ of $H$ sends $h_0\mapsto h_1$, but no inner automorphism of $H$ (conjugacy by an element of $H$) does so. For instance if $H$ is Abelian, then there are no inner automorphisms other than the identity at all. Then you can build a group $G$ which contains $H$ as a normal subgroup and in which $h,h'\in H$ are conjugate, namely the semi-direct product of $H$ and $\def\Z{\Bbb Z}\Z$, acting on $H$ by having its $1$ act as $\phi$. This is formally the set of products $hg_0^i$ for $h\in H$ and $i\in\Z$, where the group law it that suggested by the notation, together with the commutation relation $g_0h=\phi(h)g_0$. The latter ensures that $g_0h_0g_0^{-1}=h_1$.

A lot of variations to this are possible, notably on can take a cyclic group of the same order as $\phi$ in the place of$~\Z$, so that the resulting group can be finite (if $H$ is). Concrete examples where something like this happens are with $H$ a finite rotation group inside the corresponding dihedral group: every rotation is conjugate to its inverse in the dihedral group, but (except in the case where the inverse it the rotation itself) not in the rotation subgroup. Similarly every permutation of $S_n$ is conjugate to its inverse in $S_n$ (they have the same cycle structure) but usually not inside the subgroup generated by the permutation itself (and for some even permutations it may even fail inside the alternating group).

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