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Let $H, K$ be subgroups of $G$ with finite indexes, and $K\lhd G$, $H\lhd G$. Show $H \cap K$ has finite index.

We were taught only first and second homomorphisms theorems, and not all the indexes inequilities etc'. Is there a way going around these index inequilities? Also, is normality transitive?

Thanks!

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  • $\begingroup$ The claim is true also without requiring normality of $\;H,K\;$ $\endgroup$
    – DonAntonio
    Apr 20, 2014 at 15:14
  • $\begingroup$ The orbit of $H\times K$ in $G/H\times G/K$ must be finite. $\endgroup$
    – anon
    Apr 20, 2014 at 15:15

2 Answers 2

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Consider the homomorphism $\phi:G\rightarrow (G/H)\oplus (G/K)$ that sends $g$ to $(gH,gK)$. Verify that $Ker \phi=H\cap K$. By the first isomorphism theorem:

$$G/(H\cap K)\cong (G/H)\oplus (G/K)$$

Since $(G/H)\oplus (G/K)$ is finite (because $G/H$,$ G/K$ are finite), therefore $H\cap K$ has finite index

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Define a map

$$\phi: G\to G/H\times G/K\;\;,\;\;\;\phi(x):=(xH\,,\,xK)$$

The map is injective, since

$$(xH\,,\,xK)=(H\,,\,K) \iff x\in H\cap K$$

and from here that

$$G/(H\cap K)=\phi(G)\le G/H\times G/K\implies \left|G/(H\cap K)\right|\le\left|G/H\times G/K\right|$$

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