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Let $G$ be a group and let $H$ and $N$ be subgroups of $G$. Suppose that $[G:H] \leq |N|$. Does this always imply that $[G:N] \leq |H|\ $? Lagrange's theorem tells us that this is true in the finite case. What about in general?

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Umm, no. Take $G=\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}_2$, $H$ the $\mathbb{Z}_2$ factor, and $N$ infinite cyclic. –  user641 Nov 6 '11 at 19:48

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It doesn't hold in general. For example Consider $G =Z \times C_2=\langle a,b\rangle$. We take $H = \langle a^4\rangle$ and $N = \langle b\rangle$, so that $|G| = |H| = |G:N| = \aleph_0$, while $|G:H| = 4 \not\le |N| = 2$.

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Generally it is polite to credit the person who actually gave the answer, especially if you copy-paste it: John Bray in this case. –  Jack Schmidt Nov 7 '11 at 21:21

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