Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is related to this question. I just didn't want a prolonged discussion in the comments.

Let $\phi: G \to G'$ be a homomorphism. Let $G$ be a finite group. Let $K \leq G$ be the kernel of $\phi$. Let $I \leq G'$ be the image of $\phi$.

Let $H' \leq G'$. Find a formula relating the order of $\phi^{-1}(H')$ in terms of $H', I, K$.

Attempt at a solution: $|\phi^{-1}(H')|=|H'\cap I|\cdot |K \cap \phi^{-1}(H')|$.

My justification for this is as follows. $\phi^{-1}(H')/(K \cap \phi^{-1}(H'))\cong H' \cap I$

share|cite|improve this question
You wrote $\phi^{-1}(H)$ a couple of times, it should be $\phi^{-1}(H')$ always. With that fixed, it's correct, but more complicated than it need be. Since $K = \phi^{-1}(\{e\})$, you have $K \subset \phi^{-1}(H')$, so $\lvert \phi^{-1}(H')\rvert = \lvert K\rvert \cdot \lvert H'\cap I\rvert$. – Daniel Fischer Oct 31 '13 at 20:01
Thanks for that edit! – emka Oct 31 '13 at 20:01
up vote 0 down vote accepted

As long as you can justify the isomorphism in question, you should be fine (and as Daniel said, you can simplify by noting that $K\subseteq\phi^{-1}(H')$). Also, you want to make sure you're assuming that $G$ is a finite group, otherwise the formula might not make sense (if $G$ is infinite, $\left|G/H\right|\neq\left|G\right|/\left|H\right|$, because the right side is not well defined).

share|cite|improve this answer
Thanks for that. I forgot the finiteness assumption. – emka Oct 31 '13 at 20:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.