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.

Solution 1:

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.

Solution 2:

Here is another argument.

Let $\{a,b\}$ be a conjugacy class. Notice that if $aba^{-1}=a$, then $a=b$. Contradiction. So it has to be that $aba^{-1}=b$, which means that $a$ and $b$ commute.

Since $\{a,b\}$ is a conjugacy class, for any $g\in G$, $ga=ag$ or $ga=bg$. Also, $gb=ag$ or $gb=bg$.

Now, consider the subgroup $N$ generated by $a$ and $b$. $N$ is of the form $\{a^{h_1}b^{k_1}...a^{h_n}b^{k_n}|h_i, k_i\in \mathbb{Z}\}$. Since we can replace "$g$ times $a$ or $b$" by "$a$ or $b$ times $g$," $ga^{h_1}b^{k_1}...a^{h_n}b^{k_n}g^{-1}=a^{h_1'}b^{k_1'}...a^{h_m'}b^{k_m'}gg^{-1}=a^{h_1'}b^{k_1'}...a^{h_m'}b^{k_m'}$ is again in $N$. So $N$ is normal.

If $N=G$, then everything in $G$ is a product of powers of $a$ and $b$. This gives us that everything commutes with one another. So any conjugacy class can only have one element. Contradiction.

So $N$ is a proper normal subgroup.