Show that a group of order 24 with no element of order 6 is isomorphic to .

Note that , so that Sylow’s Theorem forces and .

Suppose . If , then by the recognition theorem for direct products, , where and are Sylow 2- and 3-subgroups of , respectively. By Cauchy, there exist elements and of order 2 and 3, so that has order 6, a contradiction. Suppose now that . Since mod 4, there exist such that is nontrivial and its normalizer has order , by a previous theorem. Thus is normal. Note that is either 2 or 4.

Suppose . Then at most nonidentity elements of are contained in Sylow 2-subgroups, and 3 elements are contained in Sylow 3-subgroups. This leaves 4 elements not of prime power order, one of which must have order 6- a contradiction.

Suppose now that . By the N/C Theorem, , so that is central in . By Cauchy, there exist elements and of order 2 and 3, so that has order 6, a contradiction.

Thus we may assume . Let be a Sylow 3-subgroup and let . The action of on yields a permutation representation whose kernel is contained in . Recall that normalizers of Sylow subgroups are self normalizing, so that is not normal in . Moreover, we have . We know from the classification of groups of order 6 that is isomorphic to either or ; however, in the first case we have an element of order 6, a contradiction. Thus . We know also that the normal subgroups of have order 1, 3, or 6. If , then is normal in , a contradiction. If , then by the N/C theorem we have . In particular, contains an element of order 2, so that contains an element of order 6, a contradiction.

Thus , and in fact . Since is finite, .

### Like this:

Like Loading...

*Related*

## Comments

Why are P2 and Q2 abelian ?

Good question. I’m not sure.

Okay- I think it is fixed now. I’m not sure where the abelian assumption came from.

Thanks for the tip!

At the end, n3(G)=4 forces |N|=6, no ?

You’re right.

I think it is fixed now.

Thanks again!

You’re welcome.