# Decomposition of Infinitely Generated Abelian Groups [duplicate]

I am curious if there is any structure result for an infinitely generated abelian group $$G$$?

In particular, is the following naive guess true?

$$G\cong \bigoplus_{i\in A}\mathbb{Z}\oplus \bigoplus_{i\in B }\mathbb{Z}_{q_i}$$

where $$A$$, $$B$$ are some infinite sets, and $$\mathbb{Z}_{qi}$$ are primary cyclic groups.

Or perhaps it is not a direct sum, i.e.

$$G\cong \prod_{i\in A}\mathbb{Z}\times \prod_{i\in B }\mathbb{Z}_{q_i}$$?

Thanks a lot.

## marked as duplicate by Arturo Magidin abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 16 at 4:26

• I think it's not true if you use the implied addition on the right hand side of your equation. I don't think you can write a group isomorphism between the additive group $(\mathbb{Q},+)$ and a direct sum of copies of $\mathbb{Z}$. – desiigner Jul 16 at 3:30
• @desiigner You are correct: every quotient of $\mathbb{Q}$ is finite, so we would need $\mathbb{Q}\cong X\times Y$ for some finite group $X$. However, every element of $\mathbb{Q}$ has infinite order so we cannot embed any finite group $X$ into $\mathbb{Q}$, a contradiction. – user1729 Jul 16 at 14:20