# cardinality of Galois group in $\mathbb{Q}$($\zeta_n$) [duplicate]

Let $$\mathbb{Q}$$($$\zeta_n$$) be some cyclotomic field, where $$\zeta_n$$ is a n-th root of unity.

I already managed to show that $$\mathbb{Q}$$($$\zeta_n$$) is an Galois extension, but now i struggle to show that $$[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n)$$, where $$\varphi(n)$$ is Euler's totient function.

## 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(); } ); }); }); Dec 19 '18 at 16:26

For any $$\sigma \in Gal(\mathbb Q(\zeta_n)/\mathbb Q)),$$ $$\sigma(\zeta_n)$$ must be a primitive $$n$$-th root of unity, because $$\sigma$$ is a $$\mathbb Q$$-automorphism. Since $$\zeta_n$$ generates $$\mathbb Q(\zeta_n)/\mathbb Q$$, it follows that this group has as many elements as the set of primitive $$n$$-th roots of unity, that is $$\varphi(n)$$, where $$\varphi$$ denotes Euler's totient function.