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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. $\phi^{-1}(H')/(K \cap \phi^{-1}(H'))\cong H' \cap I$

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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! –  abet Oct 31 '13 at 20:01

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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).

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Thanks for that. I forgot the finiteness assumption. –  abet Oct 31 '13 at 20:27

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