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Let $G$ be a group of order $2^n$ ($n\ge1$). Let $G'$ the derived group of $G$. Assume that $$G/G'\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$$

Can someone explain to me why $G$ admits seven normal subgroups of index 2 and seven normal subgroups index 4? Is there a general result which gives us these kinds of information about the numbers of normal subgroups of a $p$-group?

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  • $\begingroup$ How many subgroups of index $2$ does $(\Bbb Z/2\Bbb Z)^3$ have? $\endgroup$ – Kenny Lau Dec 13 '17 at 0:52
  • $\begingroup$ @KennyLau, Maybe $4$? $\endgroup$ – Med Dec 13 '17 at 0:56
  • $\begingroup$ I've added a counting argument in my answer demonstrating why it is $7$ $\endgroup$ – Kenny Lau Dec 13 '17 at 0:58
  • $\begingroup$ @KennyLau, Thanks. $\endgroup$ – Med Dec 13 '17 at 1:00
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We know that $G/G' \cong (\Bbb Z/2\Bbb Z)^3$. Let $\varphi : G \to (\Bbb Z/2\Bbb Z)^3$ be the canonical homomorphism with kernel $G'$.

Denote $(\Bbb Z/2\Bbb Z)^3$ as $E_8$. From here (or by counting), $E_8$ has $7$ subgroups of index $2$ and $7$ subgroups of index $4$, all of which are normal because $E_8$ is abelian.

Therefore, their preimages will give you the $7$ normal subgroups of index $2$ and the $7$ normal subgroups of index $4$.


Counting argument:

A subgroup of $E_8$ of index 2 are generated by $2$ elements other than the identity. Since we have $7$ elements other than the identity, this gives us $\dbinom 7 2 = 21$ choices. However, each subgroup admits $3$ choices of generators, which leaves us with $7$ subgroups of index $2$.

Any subgroup of $E_8$ of index $4$ is generated by one element, so we have $7$ choices.

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