# find a group with the property :

a)find a nontrivial group $G$ such that $G$ is isomorphic to $G \times G$

what i'm sure is that $G$ must be infinite ! but have now idea how to get or construct such group i chose many $G$'s but all of the homomorphism was not injective

b) an infinite group in which every element has finite order but for each positive integer n there is an element of order n

the group $G = (Z_1 \times Z_2 \times Z_3 \times Z_4 \times ...)$ satisfies the conditions except the one which says that every element have finite order .

how can we use this group to reach the asked group ?

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Hint for 1: Consider an infinite number of copies of the same group. Hint for 2: Remember those restricted direct products. – Tobias Kildetoft May 28 '13 at 19:34

Very nice question!

You've got some answers for $(b)$. With respect to answering $(a)$, (which is, indeed, a challenging question):

I think you'll find the following post from MathOverflow very handy:

Is there a finitely generated nontrivial group $G$ such that $G\cong G\times G$?

It was asked by Martin Brandenburg, and there are two excellent answers (in the affirmative) given there.

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Have you read the linked post, Maths Lover? – amWhy Jun 9 '13 at 22:12
+1 $\frac{+}{0}$ – Babak S. Jun 22 '13 at 0:18
@BabakS. , this is nice , + infinty O_O :D – Maths Lover Jun 22 '13 at 3:30

Hint for b: Consider a quotient of $\mathbb{Q}$.

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I think for b you can consider $G=\mathbb Z(p^{\infty})$ where $p$ is a prime as well.

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For the second problem, you can use the subgroup of $\mathbb{Z}_1\times \mathbb{Z}_2\times \cdots\times \mathbb{Z}_n \times \cdot$ consisting of all sequences $(a_1,a_2,a_3,\dots)$ such that all but finitely many of the $a_i$ are $0$.

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This subgroup is the direct sum of the $\mathbb{Z}_n$, and is often denoted by $\mathbb{Z}_1 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \ldots$ – Mikko Korhonen May 29 '13 at 8:53