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This isn't a homework problem. I'm preparing for an exam, and I have no idea how to solve this problem.

Let $G$ is a group such that every non-identity element has order $2$. Let $H$ be a subgroup. Prove $G/H$ is isomorphic to a subgroup of $G$.

The fact that every non-identity element has order $2$ means $G$ is abelian (since for all $a, b \in G, (ab)(ab) = e \implies ba = a^{-1}b^{-1}$, but $a = a^{-1}$ since $a^{2} = e$, and likewise $b= b^{-1}$, so $ba = ab$).

Since $G$ is abelian, $H$ is a normal subgroup of $G$. So $G/H$ is a group. Now what? I know that subgroups of $G/H$ are in $1-1$ correspondence with subgroups of $G$ containing $H$. Does this help?

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  • $\begingroup$ $G$ is a vector space over the two element field $F_2$; then the dimension of $G/H$ over $F_2$ is at most the dimension of $G$. $\endgroup$ – egreg Oct 24 '14 at 21:15
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Since $G$ is Abelian, I’ll write it additively. Say that $A\subseteq G$ is independent if $\sum F\ne 0_G$ for all finite $F\subseteq A$. Use Zorn’s lemma to get a maximal independent $B\subseteq G$. Show that for each $x\in G$ there is a finite $F_x\subseteq B$ such that $x=\sum F_x$; then show that $F_x$ is unique and infer that $G$ is a vector space over $\Bbb F_2$ with basis $B$.

Now modify the construction by first getting a basis $B_H$ for $H$ and then extending it to a basis $B$ for $G$, and show that $G/H$ is isomorphic to the subgroup generated by $B\setminus B_H$.

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    $\begingroup$ Thank you very much for your help. Your answer is clear and doesn't leave out any important details. You clearly know how to teach/communicate math. $\endgroup$ – layman Oct 25 '14 at 2:53
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    $\begingroup$ I agree. Wish the other 100K plus reputation would give an answer of the quality of Brian's answers. IT is always a pleasure to read Brian's answers. Hagen's answer is not really helpful. $\endgroup$ – ILoveMath Oct 25 '14 at 3:04
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    $\begingroup$ @Learner: In fairness, I would say that Hagen’s answer is good in itself; after all, it’s just a terser version of mine! But it assumes a bit more background than the OP has, so it wasn’t as helpful/useful here and now. $\endgroup$ – Brian M. Scott Oct 25 '14 at 3:25
  • $\begingroup$ @MathIsHardNoItsNot: Thank you. Much as I like mathematics, I’ve always thought of myself as a teacher first and a mathematician second, so that’s very nice to hear. $\endgroup$ – Brian M. Scott Oct 25 '14 at 3:26
  • $\begingroup$ @BrianM.Scott Why should one use Zorn's lemma to infer that an elementary $2$-group is a vector space over the two element field? $\endgroup$ – egreg Oct 26 '14 at 8:36
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$G$ is an elementary $2$-group, so it's a vector space over the field $F_2$ with two elements. Every subgroup is thus a direct summand.

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    $\begingroup$ I appreciate your time and willingness to answer my question. But your answer is too brief for me and so didn't help me very much. $\endgroup$ – layman Oct 24 '14 at 22:13
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    $\begingroup$ What is a $2$-group? How does this imply $G$ is a vector space over $\mathbb{F}_{2}$? How does this imply every subgroup is a direct summand? What is a direct summand? $\endgroup$ – layman Oct 24 '14 at 22:57
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    $\begingroup$ @MathIsHardNoItsNot, It's hard to understand how you don't know what a direct summand, a basic and rather important notion in linear algebra say, is, if you're trying to study group theory, which is usually more advanced than linear algebra. $\endgroup$ – Timbuc Oct 25 '14 at 1:19
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    $\begingroup$ @Timbuc How does this comment help me? $\endgroup$ – layman Oct 25 '14 at 1:35
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    $\begingroup$ @Timbuc This site is all about people engaging in math and discussing it with each other. Yet sometimes, and in this specific case, the level of engagement is low, and to me that is very sad. The two answers to this question look like they have been formulated for a textbook, where usually brevity is preferred because there are only finitely many pages. Well, this site isn't a textbook, and I personally don't like answers that are brief and assume the person that asked the question is already at the level to receive such short responses. If that were true, the question wouldn't be asked... $\endgroup$ – layman Oct 25 '14 at 1:55
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With a bit of linear algebra: $G$ is a vector space over the field $\mathbb F_2$ and the subspace $H$ ha a complement.

Let's stay in group theory, though: Let $K<G$ be a maximal subgroup with $K\cap H=1$ (use Zorn's lemma). Show that $G/H\approx K$

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  • $\begingroup$ I'm not sure I understand why $G$ is a vector space over $\mathbb{F}_{2}$. To be a vector space over $\mathbb{F}_{2}$, every element should be able to be written as a linear combination of some basis elements with coefficients in $\mathbb{F}_{2}$, but I don't see it... $\endgroup$ – layman Oct 24 '14 at 21:31
  • $\begingroup$ @MathIsHardNoItsNot: $G$ being Abelian, I write it additively. Say that $A\subseteq G$ is independent if $\sum F\ne 0_G$ for all finite $F\subseteq A$. Use Zorn’s lemma to get a maximal independent $B\subseteq G$. Show that for each $x\in G$ there is a finite $F_x\subseteq B$ such that $x=\sum F_x$; then show that $F_x$ is unique, and infer that $G$ is a vector space over $\Bbb F_2$ with basis $B$. Now modify the construction by first getting a basis $B_H$ for $H$ and then extending this to a basis $B$ for $G$, and show that $G/H$ is isomorphic to the subgroup generated by $B\setminus B_H$. $\endgroup$ – Brian M. Scott Oct 25 '14 at 2:36
  • $\begingroup$ @BrianM.Scott By $A \subseteq G$ being independent, do you mean no sum of elements of $A$ can be $0$? $\endgroup$ – layman Oct 25 '14 at 2:43
  • $\begingroup$ @MathIsHardNoItsNot: Almost: I defined it to mean that no finite sum can be $0_G$. $\endgroup$ – Brian M. Scott Oct 25 '14 at 2:46
  • $\begingroup$ @BrianM.Scott Ok, thank you for the clear instructions and help. I understood everything you said and will follow those steps to prove the statement. One last thing: I know it might be annoying to write everything out again, but I would like to accept your answer for this question. Do you mind reposting the information as your own answer? $\endgroup$ – layman Oct 25 '14 at 2:48

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