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I have to solve this exercise: Let $G$ a group of order $16$ such that every $g \in G$ is such that $g^2=e$. I have to determine $G$ up to isomorphisms.

We know that $G$ is abelian since $(ab)^2=e$. From the structure theorem, since the partition of $4$ are $\{4,3 1,211,1111\}$, we conclude that up to isomorphisms, $G$ is $\mathbb{Z}_{16}$, or $\mathbb{Z}_8 \times \mathbb{Z}_2$, or $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$, or $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$.

Is it right?

Shaun
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Rick88
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    Please show your attempt. – SoG Dec 14 '21 at 11:05
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    Is that the correct use of the term idempotent? I thought an idempotent element $g$ would satisfy $g^2=g$ – Arthur Vause Dec 14 '21 at 11:14
  • Yes sorry, the term is incorrect. – Rick88 Dec 14 '21 at 11:19
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    $g^2=e$ (which is not IP, but self inverse. IP would be $g^2=g$) implies that if for some subgroup $S$ we have $g\not\in S$ then $S\cap \langle g\rangle = {e}$. This implies that the subgroup induced by $S$ and $g$ can be written as $S\oplus \langle g\rangle$. Thus if $S\neq G$ you can find some $S\subset S'\subseteq G$ with $S'=S\oplus \langle g\rangle$. From this you can step by step show that $G=\langle g_1\rangle \oplus \langle g_2\rangle \oplus \langle g_3\rangle \oplus \langle g_4\rangle$ for distinct $g_i\neq e$. Then note that $\langle g_i\rangle$ is isomorphic to $\mathbb Z_2$. – Lazy Dec 14 '21 at 11:24
  • Thank you, from this the unique possibility is $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. – Rick88 Dec 14 '21 at 11:29

2 Answers2

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A group where all elements are idempotent is abelian. Therefore according to the theorem of classification of abelian groups, $G$ is isomorphic to a product $$\mathbb Z_{p_1} \oplus \cdots \oplus \mathbb Z_{p_n}$$ where $p_1, \dots, p_n$ are powers of $2$.

As all elements have order equal to $2$, $G$ is isomorphic to $\mathbb Z_{2} \oplus \mathbb Z_{2} \oplus \mathbb Z_{2} \oplus \mathbb Z_{2}$.

SoG
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mathcounterexamples.net
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First prove that $G$ must be abelian, then prove that $$(a, g) \mapsto g^a$$ gives an action $\mathbb Z/2 \times G \to G$. Finally, note that $\mathbb Z/2$ is a field.

SoG
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Jim
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