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$\newcommand{\Z}{\mathbb{Z}}$

The general afirmation would be:

Let $G$ a group, such that $g^2 = 1$ for $g \neq 1$ then

$$ G \cong \left(\frac{\Z}{2\Z}\right)^X$$.

I am trying to prove that for a special case, let $G$ with $|G| = 8$, such that $g^2 = e$ for $g \neq e$. Then

$$ G \cong \frac{\Z}{2\Z}\times\frac{\Z}{2\Z}\times\frac{\Z}{2\Z} $$

The idea is a elementary proof without the fundamental theorem of finitelly generated abelian groups.

I have proved that it is abelian and my idea is to use semidirect product with $H = \frac{\Z}{2\Z}$ and $K = G/H \cup \{e\}$ that i only miss proving is a group.

If we knew $K$ is a group $G = HK$ with $ H\cap K = e$ then it is the semidirect product, but both subgroups are abelian then are normal subgroups and would be a direct product.

Finally we will have to prove that $K \cong \frac{\Z}{2\Z}\times\frac{\Z}{2\Z}$.

Fran Aguayo
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    The dupe target goes well beyond what is needed to address the special case in the OP. That's very misleading from a didactical standpoint, which seemingly is what the OP is pursuing. Voted to reopen. – Kan't Apr 18 '24 at 18:56
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    The group hasn't got any subgroup $H\cong C_4$, so it can arise only as a semidirect product $C_2\ltimes C_2^2$. Show that if the action is nontrivial, then any such a group has an element of order $4$. Therefore the action must be trivial, and hence $G\cong C_2\times C_2^2$. – Kan't Apr 18 '24 at 19:04
  • thanks @Kan't, don't know what to change to get it reopen since the answers in the other questions all use module theory and i was looking one for the beggining of abstract algebra so a demostration with only group theory. – Fran Aguayo Apr 18 '24 at 21:33
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    Your general affirmation applies only to finite groups. There are also infinite groups with this property, and they are also abelian and direct products of copies of $C_2$. – Derek Holt Apr 19 '24 at 19:08

3 Answers3

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The following hint may help: prove that if $a^2, b^2, (ab)^2 = 1$, then $ab = ba$. From this, you have that $G$ is abelian and so are all its quotient groups. Since every non-trivial element of $G$ has order two, the same is true of elements in any quotient group. You now have a way to proceed inductively. This approach generalizes to a theorem for $|G| = 2^k$.

Robert Bell
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  • That doesnt tell us what we want $G/H \times H \cong G$ is not always true and i dont see that is true trivially in here – Fran Aguayo Apr 18 '24 at 17:12
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    I only meant this as a hint. Here's how I would finish the argument. Let $a$ be non-trivial element and let $H = {1,a}$. Then $H$ is a subgroup (since $a^2=1$) and the quotient group $G/H$ has the same properties as $G$. By induction, there is surjective homomorphism $G/H \to C_2$, where $C_2$ is cyclic of order two. Let $K$ be the kernel of the map $G \to G/H \to C_2$. Then $K$ has order $2^{k-1}$. Choose any non-trivial element $b$ in $G$ that does not belong to $K$. Then $B={1,b}$ is a subgroup, $G=BK$ and $B \cap K = {e}$. – Robert Bell Apr 20 '24 at 02:40
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Select $a\in G-\{1\}$ and $b\in G-\{a,a^2,1\}$ ($-$ is set-minus). Since $G$ is commutative and $g^2=1$ for all $g\in G$, we have $H\equiv\langle a,b \rangle=\{1,a,b,ab\}\cong \mathbb{Z}_2\times \mathbb Z_2$. For convenience set $c\equiv ab$.

Then $G=H\cup xH$ for some $x\notin H$, i.e. $$ G=\{1,a,b,c,x,xa,xb,xc \}, $$
where the eight elements above are distinct. Then $$ G=\{1,a\}\times \{1,b\}\times \{1,x\}\cong (\mathbb Z_2)^3. $$

Asigan
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Since $G$ hasn't got any cyclic subgroup of order $4$ (as otherwise it would have an element of order $4$), you are left to prove that any nontrivial action $\varphi\colon C_2\stackrel{}{\to}\operatorname{Aut}(C_2^2)$ gives rise to a semidirect product $C_2\stackrel{\varphi}{\ltimes} C_2^2$ which contains an element of order $\ne2$, and hence it is not isomorphic to our group, whence $G\cong C_2\times C_2^2$.

Let $C_2=\{1,a\}$ and $C_2^2\cong\{1,b,c,bc\}$; a nontrivial homomorphism $\varphi\colon C_2\to\operatorname{Aut}(C_2^2)\cong S_3$ must send $1$ to $\varphi_1=Id_{C_2^2}$, and $a$ to $\varphi_a$ which swaps any two amongst $b,c,bc$.

  • Case 1): $\varphi_a$ swaps $b,c$: \begin{alignat}{1} &\varphi_a(1) = 1 \\ &\varphi_a(b) = c \\ &\varphi_a(c) = b \\ &\varphi_a(bc) = bc \\ \tag1 \end{alignat} Accordingly, take for example: \begin{alignat}{1} x &:=(bc,a)*(b,1) \\ &= (bc\varphi_a(b),a) \\ &= (b,a) \\ \end{alignat} Hence: $$x^2=(b,a)*(b,a)=(b\varphi_a(b),1)=(bc,1)\ne (1,1)$$
  • Case 2): $\varphi_a$ swaps $b,bc$: \begin{alignat}{1} &\varphi_a(1) = 1 \\ &\varphi_a(b) = bc \\ &\varphi_a(c) = c \\ &\varphi_a(bc) = b \\ \tag2 \end{alignat} Accordingly, take for example: \begin{alignat}{1} x &:=(c,a)*(b,1) \\ &= (c\varphi_a(b),a) \\ &= (b,a) \\ \end{alignat} Hence, again: $$x^2=(b,a)*(b,a)=(b\varphi_a(b),1)=(c,1)\ne (1,1)$$
  • Case 3): $\varphi_a$ swaps $c,bc$: \begin{alignat}{1} &\varphi_a(1) = 1 \\ &\varphi_a(b) = b \\ &\varphi_a(c) = bc \\ &\varphi_a(bc) = c \\ \tag3 \end{alignat} Accordingly, take for example: \begin{alignat}{1} x &:=(b,a)*(c,1) \\ &= (b\varphi_a(c),a) \\ &= (c,a) \\ \end{alignat} Hence, yet again: $$x^2=(c,a)*(c,a)=(c\varphi_a(c),1)=(b,1)\ne (1,1)$$ No way, then, to get a nontrivial semidirect product $C_2\ltimes C_2^2$ with elements of order 2, only. Therefore, necessarily $G\cong C_2\times C_2^2$.
Kan't
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