# Why are there only two groups of order $21$? [duplicate]

Up to isomorphism, I would like to say that there are only two groups of order $$21$$. There is the cyclic one, and using Sylow's theorems it is easy to see that there could be at most two nonabelian ones. Is there a way to show that there is in fact only one (up to isomorphism) without having to write down all the elements and the corresponding tables (I only see that for now)?

## marked as duplicate by YuiTo Cheng, Ernie060, Dietrich Burde 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(); } ); }); }); May 7 at 12:38

The Sylow theorems yield that a group of order $$21=3*7$$ has either $$1$$ or $$7$$ Sylow- $$3$$-subgroups and either $$1$$ or $$8$$ Sylow-$$7$$-subgroups.

Now, $$7$$ and $$8$$ can't occur simultaneously by a simple counting argument.

Also any subgroup of order $$7$$ is normal because its index is the smallest prime dividing $$21$$. Thus in fact there is a unique Sylow $$7$$-subgroup.

So, if $$G$$ is non-abelian we have $$1$$ unique, normal Sylow $$7$$-group $$P_7$$ and $$7$$ Sylow-$$3$$-subgroups.

It follows that $$G\cong P_7 \rtimes_\phi P_3$$ with $$\phi: P_3 \to Aut(P_7)$$ a morphism.

Suppose a generator of $$Aut(P_7) \cong C_6$$ is $$s$$ and a generator of $$P_3$$ is $$t$$. The morphism $$\phi$$ is non-trivial, and thus must send the generator $$t$$ to an element of order $$3$$. Thus it must get mapped to either $$s^2$$ or $$s^4$$. Thus, there are at most two morphisms (and in fact precisely two) possibilities for $$\phi$$. Denote them with $$\phi_1$$ and $$\phi_2$$. Denote inversion in $$P_3$$ with $$f$$. Then we have $$\phi_1 = \phi_2 \circ f$$ and thus because $$f\in Aut(P_3)$$ we have $$G\cong P_7 \rtimes_{\phi_1} P_3 \cong P_7 \rtimes_{\phi_2} P_3$$. (--- Here I used that we have $$N\rtimes_\phi K \cong N \rtimes_{\phi \circ f} K$$ where $$f \in Aut(K)$$. An explicit isomorphism between the two is not too hard to define---). This reasoning shows that there are at most, and in fact, again, precisely, two isomorphic types of $$G$$. One is abelian, the other one is not, and can be obtained by taking a non-trivial morphism $$C_3 \to Aut(C_7)$$ and forming the associated semidirect product.

• That is my problem: isn't the semi direct product defined by a morphism $\phi : P_3 \to Aut(P_7)$, so that this would mean looking at the image of a generator that the possible $\phi$ are determined by the ordre 3 elements in $Aut(P_7) \cong P_6$. There are however two such elements, isn't it? So that adding the identity (i.e. the direct product, that is the cyclic group $P_{21}$), that would give three groups. We still need to show that the two non direct semi direct products are isomorphic, and this is exactly my question :) – Wolker May 7 at 7:37
• If $y$ is a generator of $P_3$, then replacing $y$ with $y^{-1}$ results in the other semidirect product, so they are isomorphic. – Derek Holt May 7 at 7:41
• As promised, I added some details. Can you let me know if everything is clear? I am also new to this material and would love some feedback :). – EpsilonDelta May 8 at 17:21