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Consider the finite 2-groups containing cyclic subgroup of index 2:

$C_{2^n}$, $C_{2^{n-1}}\times C_2$, $D_{2^n}$, $SD_{2^n}$, $QD_{2^n}$, $Q_{2^n}$.

Can every finite (non-abelian) 2-group be embedded in $H\times K$ where $H$, and $K$ are one of the groups in the list?

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  • $\begingroup$ It isn't necessary to put numbers inside dollar signs, and it slows down the page loading (and if MathJax is on the blink it's ugly), so I cleaned your post a little. $\endgroup$
    – David Roberts ♦
    Commented Sep 7, 2011 at 5:25

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Any subgroup of $H \times K$ will have a subgroup of index at most 4 that is a product of two cyclic groups. This comes from intersecting with a fixed choice of cyclic subgroups of index 2 in each factor.

The direct sum of 3 copies of $\mathbb{Z}/8\mathbb{Z}$ is a counterexample, since it doesn't have a subgroup of index at most 4 that is a product of 2 cyclic groups. If you want a nonabelian group, you can take the product of this with quaternions.

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