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

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?

share|cite|improve this question
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
up vote 5 down vote accepted

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 $|G:H| = 4 \not\le |N| = 2$.

share|cite|improve this answer
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

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.