# What finite groups have unique cyclic subgroups? [duplicate]

Let $G$ be a finite group of order $n$. If $G$ is cyclic, we then know that are subgroups are cyclic are are unique. If $G=\langle x\rangle$, and $d|n$, then $\langle x^d\rangle$ is the unique cyclic subgroup of order $\frac{n}{d}$.

However, suppose we know that the group $G$ has a unique subgroup of order $d$ for any $d$ such that $d|n$. What else can we say about $G$? Does it have to be cyclic? Can it be factorizable over subgroup $H$ and $K$?

Any help is greatly appreciated!

## marked as duplicate by Derek Holt finite-groups 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(); } ); }); }); May 12 '17 at 16:20

This condition implies that the Sylow $p$-subgroup of $p$ is unique and cyclic. This implies that $G$ is the direct product of its Sylow subgroups, so is cyclic.