I've stumbled upon this exercise and it's in a chapter on Homomorphisms before Quotient groups are introduced, so I am interested in a solution that does not use them.

Here's the full exercise:

Let $G$ be a finite group and $H$ be a subgroup of $G$ and $N(H)$ the normalizer of $H$.

Let $K$ be any subgroup of $G$ and $K^* = \{N(H)k : k \in K\}$, $X_K = \{kHk^{-1} : k \in K\}$

  1. Prove that $X_K$ is in one-to-one correspondence with $K^*$.
  2. Conclude that the number of elements in $X_K$ is a divisor of $|K|$.

I've done 1) which comes down to the fact that if $N(H)k_1=N(H)k_2 \Leftrightarrow k_1Hk_1^{-1}=k_2Hk_2^{-1}$.

However I do not know how to proceed with 2). I can probably use Quotient groups, but I am thinking there should be a simpler solution without using them since they are introduced in the next chapter.

  • 2
    $\begingroup$ Presumably you do have Lagrange's theorem. If so, then show that $|K^*| = |\{(K \cap N(H))k : k \in K\}|$, giving $|K^*| = [K : K \cap N(H)]$, from which the result follows by two applications of Lagrange. $\endgroup$
    – Rob Arthan
    Oct 12, 2020 at 22:04
  • $\begingroup$ @RobArthan would accept your comment as an answer. Though I am impressed that the exercise requires this step, since the previous ones were straightforward, ie. just an application of a definition/etc. So I guess I was thinking the solution had to be trivial application of the definitions but it's not. $\endgroup$ Oct 13, 2020 at 21:20
  • $\begingroup$ I've posted my comment as an answer (with a slight generalisation of the reasoning). It sounds like you textbook is interleaving routine exercises with one's that require a little bit more investigation. $\endgroup$
    – Rob Arthan
    Oct 13, 2020 at 21:38
  • $\begingroup$ Well, so far I've seen some difficult exercises in Pinter, but the lack of an indication is a tad bit annoying, I mean there could be a symbol near the exercise number if not a solution. But here it says: "Conclude...", as if 2) is a direct corollary of 1). $\endgroup$ Oct 13, 2020 at 21:41

1 Answer 1


If $J$ is any subgroup of $G$, then for $k_1, k_2 \in K$, $Jk_1 = Jk_2$ iff $k_1k_2^{-1} \in J$ iff $k_1k_2^{-1} \in J \cap K$ iff $(K \cap J)k_1 = (K \cap J)k_2$. So $|\{Jk : k \in K\}| = |\{(K \cap J)k : k \in K\}| = [K : K \cap J]$. By Langrange's theorem, this tells us that $|\{Jk : k \in K\}|$ divides $|K|$. Taking $J=N(H)$ and using part 1 of the problem, this gives us that $|X_K| = |K^*|$ divides $|K|$.

  • $\begingroup$ We do not need to prove that $|K^*|$ divides |G|, just |K|, but this is implied I guess. $\endgroup$ Oct 13, 2020 at 21:38
  • 1
    $\begingroup$ Sorry. I misread your question. I will edit. $\endgroup$
    – Rob Arthan
    Oct 13, 2020 at 21:39

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