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Let $G$ be a finite group such that $x^2 = e$ for each $x \in G$. I know already that $G$ is abelian and that the order of $G$ is $|G| = 2^n$ for some $n \geq 0$. Now I wish to show that $$G \cong \mathbb{Z}_2 \times ... \times \mathbb{Z}_2$$ where there are $n$ factors (we define the direct product of zero groups as the trivial group $\{e\}$).

So far I have the following partial proof:

Using the fact that $|G| = 2^n$, we prove by induction on $n$. When $n = 0$ we have $|G| = 2^0 = 1$, hence $G \cong \{e\}$, which we said above is isomorphic to the direct product of zero copies of $\mathbb{Z}_2$.

Now suppose for some $n \geq 1$ that the statement holds for all $k$ with $0 \leq k \leq n - 1$. We wish to prove the statement for $n$.

Since $n \geq 1$, the order of $G$ is $|G| = 2^n \geq 2$, hence $G$ is nontrivial. Therefore there is an element $x \in G$ such that $x \neq e$. The element $x$ generates a $2$ element subgroup $H_1 = \langle x \rangle$ which is isomorphic to $\mathbb{Z}_2$. Given any coset $xH_1 \in G/H_1$, we have $$(xH_1)^2 = (x^2)H_1 = eH_1$$ (Why? Or is this false?). Therefore we may apply the inductive hypothesis, i.e., $$G/H_1 \cong (\mathbb{Z}_2)^{n-1}$$ where $(\mathbb{Z}_2)^{n-1}$ is the direct product of $n-1$ copies of $\mathbb{Z}_2$.

What remains to be shown is that there is a subgroup $H_2$ of $G$ isomorphic to $G/H_1$, and that $H_1 \cap H_2 = \{e\}$ and $H_1 \cdot H_2 = G$. If we can show this then $G \cong H_1 \times H_2$ since both $H_1$ and $H_2$ are normal in the abelian group $G$. Then we may just substitute our isomorphic values of $H_1$ and $H_2$ to get the result.

agcha
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  • It helps if you first show that $G$ has to be abelian: $(xy)^2=xyxy=e=x^2y^2$ ... – Nicky Hekster Feb 08 '16 at 21:58
  • @NickyHekster the OP has done that - see the first line – Mathmo123 Feb 08 '16 at 22:04
  • The first answer to this slightly different question is very relevant http://math.stackexchange.com/questions/17054/group-where-every-element-is-order-2 – Mathmo123 Feb 08 '16 at 22:22
  • @agcha Have you been able to solve this? I am looking at the same problem. How do you know $|G|=2^{n}$? Is there a way to do this proof without cosets? (Simple concept, but I haven't been introduced to them yet and would prefer a more elementary approach as cumbersome as it may be.) Perhaps there is a subgroup of $G$ which has order $2^{n-1}$ and we can apply inductive hypothesis to that and use your theorem together with $H_1$? – Jacopo Stifani Feb 18 '16 at 02:44
  • @JacopoStifani Note that since $x^2 = e$ for each $x \in G$, every element of $G$ has order at most $2$. If G is the trivial group then $|G|=2^0$. Otherwise, there is an element $g \in G$ such that $g \neq e$, hence $g$ has order exactly $2$. The element $g$ generates a $2$ element subgroup, hence by Lagrange's theorem $2$ divides $|G|$. Suppose there is a prime divisor $p$ of $|G|$ such that $p \neq 2$. The number $1$ isn't prime, so $p > 2$, and hence by Cauchy's theorem there is an element $y \in G$ with order $p > 2$, a contradiction. – agcha Feb 19 '16 at 18:51
  • @JacopoStifani By the existence of prime factorization, we can express $|G|$ as the product of primes. We showed that $2$ is the only prime divisor of $|G|$, hence such a factorization must take the form $|G| = 2^n$ for some $n$. – agcha Feb 19 '16 at 18:55
  • @agcha Makes sense now. Thank you. – Jacopo Stifani Feb 20 '16 at 04:52

2 Answers2

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By the Fundamental Theorem of finitely generated Abelian groups our group is a direct sum of cyclic groups. But all elements have order $\le 2$. So each of the summands is isomorphic to $\mathbb{Z}_2$.

André Nicolas
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  • Thanks. Can this be done without the Fundamental theorem using more basic methods, i.e. following my attempt at the proof? – agcha Feb 08 '16 at 22:28
  • Undoubtedly, since it uses a very special case of the Theorem. The path you were attempting has somewhat the feel of a proof of the Theorem. Maybe it can be pushed through. – André Nicolas Feb 08 '16 at 22:32
  • @agha: Note that the solution suggested by egreg is very simple. – André Nicolas Feb 09 '16 at 05:42
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$G$ is in a natural way a vector space over the two element field.

egreg
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