# It is true that every group that has a finite number of subgroups is finite? [duplicate]

It is true that every group that has a finite number of subgroups is finite?

I think not, but I can not find counterexamples.

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## marked as duplicate by Martin Brandenburg 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 }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 21 '14 at 13:31

– Jérémy Blanc Jun 21 '14 at 13:21

Yes, true - look at $\langle g\rangle$ for every $g \in G$ and note that an infinite cyclic group has an infinite number of subgroups.
$\langle g \rangle$ must be finite, and since $G$ has only a finite number of subgroups, we get that $G=\bigcup_{g \in G}\langle g \rangle$ is a finite union. hence $G$ must be finite after all!
This part of that an infinite cyclic group has an infinite number of subgroups is because each $g^n$ would create a new subgroup $<g^n>$? – Croos Jun 21 '14 at 13:53