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Suppose $A \subset C$ and $B \subset C$. Assume $A$ and $B$ are conjugate subgroups, that is $cAc^{-1}=B$ for some $c \in C$. Is the following statement true?

$A=B$ if and only if $A$ and $B$ are normal subgroups of $C$

Question: If we assume $A$ and $B$ are conjugate subgroups of $C$, are they normal subgroups?

The statement above is a statement that I come out on my own to aid me to understand relationship between regular covering and conjugacy class.

EDIT: How to relate conjugacy class with regular covering/ normal covering?

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    $\begingroup$ If $cAc^{-1} = B$ for all $c\in C$, then $A = B$ by setting $c$ to be the group identity. It then follows that $cAc^{-1} = A$ for all $c\in C$, whence $A$ is normal in $C$. $\endgroup$ – Prahlad Vaidyanathan May 3 '14 at 8:26
  • $\begingroup$ Actually I know how to prove the statement above. My only confusion is when we say two groups are conjugate, do we need the assumption that both of them are normal subgroups of a group? $\endgroup$ – Idonknow May 3 '14 at 8:28
  • $\begingroup$ @JyrkiLahtonen: Thx for pointing out the mistake $\endgroup$ – Idonknow May 3 '14 at 8:53
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Fixing that one error hopefully means that you are well on your way to understanding the subtleties (conjugation, normalcy) here. It is very much possible that $cAc^{-1}=A$ (or $A=B$ if you wish) for some but not all $c\in C$. The latter is equivalent to $A$ being normal in $C$, $A\unlhd C$.

The in-between cases can be described using what's called the normalizer of $A$ in $C$. This is the set $$ N_C(A)=\{c\in C\mid cAc^{-1}=A\}. $$ It is a standard application of the subgroup criterion to show that $N_C(A)$ is a subgroup of $C$. Leaving that and the following facts as exercises:

  • We always have $A\unlhd N_C(A)\le C$.
  • $A$ is a normal subgroup of $C$ if and only if $N_C(A)=C$.
  • The normalizer $N_C(A)$ is the largest subgroup $H\le C$ with the property $A\unlhd H$.

As an example let us look at the case $C=S_4$, and $A=\langle(12)\rangle$ - the cyclic subgroup of order generated by the 2-cycle $(12)$. Here it is not too hard to show that (depends on how familiar you with the algebra of permutations) $$ N_C(A)=\{1_C, (12), (34), (12)(34)\}. $$ For example with $c=(34)$ you get $cAc^{-1}=A$ because $(12)$ and $(34)$ commute. In this case $N_C(A)$ is properly between $A$ and $C$. When we pick $c$ outside $N_C(A)$, say $c=(23)$, we get $$ B=cAc^{-1}=\langle(13)\rangle\neq A. $$

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