If a group $G$ contains an element a having exactly two conjugates, then $G$ has a proper normal subgroup $N \ne {e}$

If a group $G$ contains an element a having exactly two conjugates, then $G$ has a proper normal subgroup $N \ne {e}$

So my take on this is as follows: If we take $C_G(S)$ of S. This is a subgroup of G. If $C_G(S)=G$, then S has no conjugate but itself, so therefore $C_G(S)$ is a proper subgroup. If we suppose $C_G(S)={e}$,then in order for there to be exactly two conjugates of S, then

For every $a \ne b \in G \ {e}, bxb^{-1} =axa^{-1}$ but $bxb^{-1}=axa^{-1} \to (a^{-1}b)xb^{-1}=xa^{-1} \to (a^{-1}b)x(a^{-1}b)^{-1}=x \to a^{-1}b \in C_G(S)$

Which means that $C_G(S)$ is actually nontrivial or that $a^{-1}b=e$ if and only if $a=b$, which would be a contradiction. Thus $C_G(S)$ is a nontrivial proper subgroup. Since there are exactly 2 conjugacy classes of S and they are in one to one correspondence with cosets of S, its centralizers' index $[G:C_G(S)]=s$. Subgroups of index $2$ are normal, so $C_G(S)$ is a proper nontrivial normal subgroup.

This approach seemed very different from other examples I have seen so I guess I am wondering if this approach makes sense.

• The basic idea is correct: there is a subgroup of index $2$, and subgroups of index $2$ are always normal. However, I find your explanation a little confusing in places. The subgroup of index $2$ is the centralizer of an element $x$ such that $x$ has precisely two conjugates in $G$. – Geoff Robinson Mar 11 '15 at 1:23
• Instead of considering $C_G(S)$, streamline it by considering $C_G(x)$ where $x$ is the element of $G$ possessing exactly two conjugates. – David Wheeler Mar 11 '15 at 1:25
• Shouldn't the order of $G$ be $2<|G|<\infty$? – big-lion May 27 at 2:57

Let $$a$$ be an element with exactly two conjugates, itself and some other element $$b$$.

Let $$G$$ act on the $$X = \{a,b\}$$ by conjugation. Consider the class equation for this action: $$|X| = |X_0| + \sum |G/G_x| = 2.$$

We notice that $$|X_0| = 1$$ (since the identity actions lies in $$X_0$$, and $$a$$ isn't fixed by the action of $$b$$). Hence there is a unique element $$x\in X$$ such that $$[G:G_x] = 2$$, and then $$G_x$$ must be normal since every subgroup of index 2 is normal.

Simplifying $$a,b$$ argument a little:

Let $$g$$ be the element with exactly two conjugates. Suppose $$C_G(g) = \{e\}$$. Since $$g \in C_G(g)$$, this means $$g = e$$.

By Lagrange's Theorem, $$[G : C_G(g)] = 2$$ imples $$\lvert G \rvert = 2$$, so $$G = \{e, h\}$$. The two conjugates of $$g$$ are itself and $$hgh^{-1}$$, but both are $$e$$, contradicting $$g$$ having two conjugates.

Conclude $$C_G(g) \ne \{e\}$$.