# Prove that $|\{N(H)k : k \in K\}|$ divides $|K|$ for $H$,$K$ subgroups of a finite group $G$ without using Quotient groups

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.

• 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. Oct 12, 2020 at 22:04
• @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. Oct 13, 2020 at 21:20
• 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. Oct 13, 2020 at 21:38
• 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). Oct 13, 2020 at 21:41

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 $$|K|$$.
• We do not need to prove that $|K^*|$ divides |G|, just |K|, but this is implied I guess. Oct 13, 2020 at 21:38