# Is it true that every cyclic group of order $n$ contains exactly the same number of subgroups as the number of divisors of $n$? [duplicate]

The fundamental theorem of cyclic groups says that a for a cyclic group $$G$$ where $$|G|=n$$, for every divisor $$d$$ of $$n$$, there is a unique subgroup of $$G$$ with order $$d$$. My question is that whether these are the only subgroups of any cyclic groups? In other words, is it true that $$G$$ contains exactly $$m$$ subgroups where $$m$$ is the number of divisors of $$|G|$$?

## marked as duplicate by Dietrich Burde 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(); } ); }); }); Oct 22 '18 at 20:49

Yes. You know that for each divisor $$d$$ of $$n$$ there is a unique subgroup of order $$d$$. And Lagrangre's theorem tells us that if $$k$$ is not a divisor of $$n$$ then there can't be a subgroup of order $$k$$. Hence there can't be subgroups of other orders.