is it true the following statement:
Let $G$ be a group and let $H$ be a subgroup of $G$. If the index $[G:H]$ of $H$ in $G$ is finite, then $H$ have finitely many conjugates.
What I think, is that it's true. I proceed as follows:
Suppose $[G:H]=n<\infty$ and let $\{g_{1}H,\dots,g_{n}H\}$ be its left cosets. For every $g\in G$ there exists a unique $1\leq i\leq n$ such that $g=g_{i}h$ for some $h\in H$. This implies that $gHg^{-1}=g_{i}Hg_{i}^{-1}$. So there can be at most $n$ conjugates of $H$.
Is this correct? Thanks in advance.
Best Answer
The proof is correct, up to a minor detail.
You know that $gH=g_iH$, so $$ gHg^{-1}=g_iHg^{-1} $$ However, $g=g_ih$ for some $h\in H$, so $g^{-1}=h^{-1}g_i^{-1}$, so $$ gHg^{-1}=g_iHg^{-1}=g_iHh^{-1}g_i^{-1}=g_iHg_i^{-1} $$
This is, however, just the observation that $xH=yH$ if and only if $Hx^{-1}=Hy^{-1}$, so maybe you had already this implicit in your proof.