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Question: if $G=\mathbb{Z_2}\oplus\mathbb{Z_2}\oplus...\oplus\mathbb{Z_2}$($n$ copies where $n≥3$) then number of subgroups of $G$ which are isomorphic to $\mathbb{Z_2}\oplus\mathbb{Z_2}$ is ?

My attempt: when $G=\mathbb{Z_2}\oplus\mathbb{Z_2}\oplus\mathbb{Z_2}$ I saw there are $7$ distinct subgroups which are isomorphic to $\mathbb{Z_2}\oplus\mathbb{Z_2}$. But for this, I calculated each of them by hand.They are $<(0,0,1),(0,1,0)>, <(0,0,1),(1,0,0)>, <(0,0,1),(1,1,0)>, <(0,1,0),(1,0,0)>, <(0,1,0),(1,1,1)>, <(1,1,0),(0,1,1)>, <(1,1,1),(0,1,1)>$ where $<a,b>$ means subgroup generated by $a$ and $b$.

Now when $n≥4$ how to find number of such subgroups? and which are they? Please help

Akash Patalwanshi
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  • To avoid a wrong impression: I didn't answer because of but rather despite the request "Please help" in boldface. – joriki Mar 07 '20 at 05:51

1 Answers1

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Any two distinct non-zero elements of $G$ generate a subgroup isomorphic to $\mathbb Z_2\oplus\mathbb Z_2$. Each such subgroup is generated by any pair of its non-zero elements. Thus there are

$$ \frac{\binom{2^n-1}2}{\binom32}=\frac{\left(2^n-1\right)\left(2^n-2\right)}6 $$

such subgroups.

joriki
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  • Sir is there is any proof to your formula. Please elaborate – Akash Patalwanshi Mar 07 '20 at 06:02
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    @AkashPatalwanshi: Sure, the proof is in the sentence above. There are $\binom{2^n-1}2$ different pairs that generate these subgroups, and each subgroup is generated by $\binom32$ pairs; hence there are $\frac{\binom{2^n-1}2}{\binom32}$ subgroups. – joriki Mar 07 '20 at 06:04
  • sir. One more thing I like to ask each subgroup is generated by $\binom32$ pairs? Didn't get it – Akash Patalwanshi Mar 07 '20 at 06:25
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    @AkashPatalwanshi: Each subgroup has $2\cdot2=4$ elements, of which $3$ are non-zero. There are $\binom32$ ways to choose pairs from $3$ elements. – joriki Mar 07 '20 at 06:28