If a subgroup $H$ has index $n$ in $G$ does that mean that $H$ has n conjugacy classes when partitioned by elements of $G$?

Question

If a subgroup $H$ has index $n$ in $G$ does that mean that $H$ has $n$ conjugacy classes when partitioned by elements of $G$?

I think this is probably a weird question, but I just think the index of $H$ being $n$ means that the elements of $G$ partition something like a collection of cosets: $\{g_1H,......,g_nH\}$, is this set identical to $\{g_1Hg_1^{-1},......,g_nHg_n^{-1}\}$??

Thanks!

Answer 1

Asserting that $H$ has index $n$ in $G$ is the same thing as asserting that the set $\{gH\mid g\in G\}$ has $n$ elements. In other wrods, there are $n$ elements $g_1,\ldots,g_n\in G$ such that$$\{gH\mid g\in G\}=\{g_1H,\ldots,g_nH\}.$$But, in general, this set is not $\{g_1H{g_1}^{-1},\ldots,g_nH{g_n}^{-1}\}$. If $H$ is a normal subgroup of $G$, then$$\{g_1H{g_1}^{-1},\ldots,g_nH{g_n}^{-1}\}=\{H\},$$which consists of a single element.