# let $G$ be an infinite group of the form $G_1 \oplus G_2 \oplus \dots \oplus G_n$ , $n>1$. Prove that $G$ is not cyclic

let $G$ be an infinite group of the form $G_1 \oplus G_2 \oplus \dots \oplus G_n$ where each $G_i$ is a non trivial group and $n>1$. Prove that $G$ is not cyclic.

Attempt : Let $G = G_1 \oplus G_2 \oplus \dots \oplus G_n$ be cyclic.

then $\exists ~g =(g_1,g_2,......,g_n)$ such that $g$ is generator of $G$ if and only if :

$(i)~~ g_1$ generates $G_1, ~g_2$ generates $G_2$, and so on.

$(ii)~\gcd(|g_i|,|g_j| =1)$ whenever $i \neq j$ .

Since a group of prime order is cyclic, this means that if we take $G_i$ such that $|G_i|=p_i$ , where $p_i$ is a prime number and for any $i \neq j,~\gcd (p_i,p_j) =1$ , hence, if we take $G_i$'s in this fashion such that $g_i$ of prime order generates $G_i$, $G$ should turn out to be cylic?

Where could I be making a mistake? Thank you for the help.

• What? Surely you mean a nontrivial group. – user98602 Apr 20 '14 at 3:41
• Yes, I corrected that – MathMan Apr 20 '14 at 3:44
• The error is in taking gcds: there's no guarantee that $|g_i|$ is always finite. In fact, it can't be; since $G$ is infinite, one of the $G_i$ is going to have to be infinite. – user98602 Apr 20 '14 at 3:48
• So, since, one of the $|g_i|$ tends to $\infty$, hence, we can't be sure that $gcd(g_i,g_j) = 1$ . Hence,$G$ can't be cyclic? Is that it? – MathMan Apr 20 '14 at 3:53
• I don't understand what you mean by "$|g_i|$ tends to $\infty$". It is $\infty$, and thus $\gcd(|g_i|, |g_j|)=\gcd(\infty, |g_j|)$ is nonsense. – user98602 Apr 20 '14 at 3:57

You can prove this as follows: Suppose $G$ is cyclic, then each $G_i$ must be cyclic (being subgroups of $G$). Since $G$ is infinite, one $G_i$ must be infinite; so assume without loss of generality that $G_1$ is infinite and cyclic, so $G_1 \cong \mathbb{Z}$.
Choosing $G_2$ to be infinite ($\cong \mathbb{Z}$) or finite ($\cong \mathbb{Z}/n\mathbb{Z}$) gives two cases :
In order to show $G$ is not cyclic, you now reduce to showing that both $\mathbb{Z}\oplus \mathbb{Z}$ and $\mathbb{Z}\oplus \mathbb{Z}/n\mathbb{Z}$ are not cyclic. Each of these is fairly simple; so you get a contradiction.
Since the only infinite cyclic group, up to isomorphism, is the additive group $\def\Z{\Bbb Z}\Z$, this amounts to showing that $\Z$ cannot be written as direct sum of more than one nontrivial group. As $\Z$ has no torsion, such a direct sum would have to be a sum of infinite subgroups, but this is easily seen to be impossible. For instance the quotient by one of the factors would be isomorphic to the sum of the remaining factors, hence infinite, but any quotient of $\Z$ by a nontrivial subgroup is known to finite.