# Weaker form of inverse Galois problem [duplicate]

I do not know if this correct. So I ask.

Given a number $n\geq 2$, can we find a Galois extension of $\mathbb Q$ such that the group has order $n$? Similarly, given $n\in \mathbb N$ can we find a totally imaginary number field that is a Galois extension of $\mathbb Q$ and for which the order of its Galois group is $2n$?

## marked as duplicate by Pierre-Guy Plamondon, Namaste group-theory 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(); } ); }); }); Dec 2 '17 at 1:45

• According to this question and answer, there is always a real and cyclic extension of $\mathbb{Q}$ of degree $n$ (for any $n\geq 2$). To get a totally imaginary extension of degree $2n$, simply add a square root of $-1$. – Pierre-Guy Plamondon Dec 1 '17 at 12:55