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Suppose $H$ and $K$ are subgroups of finite index of a group $G$ (which maybe infinite), $|G:H|=m$ and $|G:K|=n$ prove that $\operatorname{l.c.m.}(m,n)\leq|G:H\cap K|\leq mn$. I am not sure about my proof for the second inequality is right or not. I do not know how to show first inequality when $|G|$ is infinite.

Note $H\cap K\leq G$. So $|G:H\cap K|=\dfrac{|G|}{|H\cap K|}=\dfrac{|G||HK|}{|H| |K|}\leq\dfrac{|G||G|}{|H| |K|}=mn$ by $|HK|\leq|G|$ as $HK\subseteq G$. Now I have no idea how to show the first inequality if G is infinite. I even do not know how to start it if $G$ is finite. I am not sure about the above proof above working for $|G|$ is infinite which seems to be the case in my perspective. Any hint will be helpful.

Michael Hardy
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user45765
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1 Answers1

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The issue with the calculational approach you present is that the equalities and divisions involved are dubious if the cardinal numbers are infinite. How should we interpret them? Instead,

Hint (1): there is an injection $G/(H\cap K)\to G/H\times G/K$, which is well-defined even if none of the denominators present are normal subgroups. (Interpreting coset spaces as "$G$-sets," instead of being a group homomorphism, this is instead a homomorphism of $G$-sets, aka "equivariant.")

Hint (2): by property [blank], we know $[G:H\cap K]$ is divisible by [blank] and [blank].

anon
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