# Groups of order 12 whose 3-Sylow subgroups are not normal.

1. Mar 23, 2013

### Artusartos

There is a corollary in our textbook that states "Let G be a group of order 12 whose 3-Sylow subgroups are not normal. THen G is isomorphic to $A_4$."

I attached the proof of this corollary and an additional corollary and proposition that was used for the proof.

The 2nd last paragraph is a bit confusing for me...

"As shown in either Problem 14 of Exercises 4.6.13 or Problem 3 of Exercises 4.7.27,
the automorphism group of $Z_2 × Z_2$ is isomorphic to S3. It’s easy to see that there are only two nontrivial homomorphisms from $Z_3$ into $S_3 \cong D_6$, and that one is obtained from the other by precomposing with an automorphism of $Z_3$. Thus, Proposition 4.6.11 shows that there is exactly one isomorphism class of semidirect products $(Z_2×Z_2) \rtimes_{\alpha} Z_3$ which is nonabelian."

1) I'm not sure how it is "easy to see that there are only two nontrivial homomorphisms"...and I don't understand what the phrase "and that one is obtained from the other by precomposing with an automorphism of $Z_3$" means. What do "one" and "other" refer to?

2) "Thus, Proposition 4.6.11 shows that there is exactly one isomorphism class of semidirect products $(Z_2×Z_2) \rtimes_{\alpha} Z_3$ which is nonabelian." I'm not sure how the proposition allows us to see that.

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• ###### Steinberger 144.jpg
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2. Mar 23, 2013

### wisvuze

A homomorphism of Z_3 into S_3 must be an injective homomorphism, as Z_3 does not contain any non-trivial subgroups [ suppose f is a homomorphism from Z_3 into S_3, then |im( f )| = | Z_3 / ker(f ) | i.e. the image of f must divide |Z_3| = 3. So it is either injective, or trivial ].
Now, in general for finite groups, any injective homomorphism f from groups G, H can be written as f compose g where g is an automorphism of G. To see this, just note that given any injective homomorphism f, you can produce a new injective homomorphism by pre-composing with an automorphism. Conversely, given 2 distinct injective homomorphisms h , k ( k^-1 compose h ) = T is an automorphism of G such that h = k compose T.
So, in Z_3 , there are only 2 automorphisms, that is z -> z^2 and z -> z ( identity ).
So, there are only 2 possible injective homomorphisms in S_3.
Thus, our two possible homomorphisms from Z_3 into S_3 can be written as phi and phi compose g where g is an automorphism of Z_3. I.e. we can write f for one homomorphism into S_3, then another is f ' = f compose the h: z -> z^2 map. So, we have the relation f = f ' compose g^-1. In particular, f and f ' compose g^-1 are conjugate ( equal ), so you can apply condition 1 in proposition 4.6.11 to say that the resulting semi-direct products are isomorphic