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Suppose $G$ is a finite Abelian group and G has no elements of order 2. Show that the map $\phi: G \to G$ given by $g \to g^2$ is an isomorphism. Is the map an isomorphism if we do not assume $G$ is finite?

Attempt:
Take $x,y \in G$. Then $\phi(xy) = (xy)^2=xyxy=x^2y^2=\phi(x)\phi(y)$. Hence $\phi$ is homomorphic.

Assume $\phi(x) = \phi(y) \implies x^2 = y^2 \implies x^2y^{-2} = e \implies (xy^{-1})^2=e$. Since there is no element of order 2, then $xy^{-1}$ must be identity. This means $x = y$. Hence injective.

I am having trouble proving that the map is surjective.
Also, what happens if we do not assume $G$ is finite? (I'd assume it won't be an isomorphism anymore, but why?)

Oran
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1 Answers1

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Finite set $X$ has this nice property that a function $X\to X$ is injective if and only if it is surjective (see answers to this: Surjectivity implies injectivity and conversely). Thus for finite groups it is automatically surjective if injective.

In the infinite case our $\phi$ does not have to be surjective (although it is still injective under "no element of order $2$" assumption). A simple example is $G=\mathbb{Z}$ the group of integers with standard addition.

freakish
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  • That is a nice property that I will check out. But also, could you please explain the intuition behind why it doesn't hold for infinite case? – Oran Jun 10 '22 at 06:15
  • +1 Just to preemptively point out the trap for n00bs: in the simple example, the operation is addition, so "$x^2$" really means $x + x = 2x$. The range is all the even integers! – Theo Bendit Jun 10 '22 at 06:18
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    @Oran I gave you an example. For $\mathbb{Z}$ our function is simply $n\mapsto 2n$. It is injective but not surjective. – freakish Jun 10 '22 at 06:18
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    To gain intuition for why "injectivity $\Leftrightarrow$ surjectivity" fails for infinite sets, look up Hilbert's hotel – SolubleFish Jun 10 '22 at 06:20
  • Haha fair enough. Thank you. – Oran Jun 10 '22 at 06:21