Show G/\Phi(G) is Elementary Abelian p-Group

[math_grl]

Homework Statement

G is a finite p-group, show that $$G/ \Phi (G)$$ is elementary abelian p-group.

Homework Equations

$$\Phi (G)$$ is the intersection of all maximal subgroups of G.

The Attempt at a Solution

By sylow's theorem's we have 1 Sylow p-subgroup which is normal, call P. Then the order of G/P = p, so it's cyclic and thus abelian. Since there is only one maximal subgroup then $$\Phi (G) = P$$?.

I'm having trouble convincing myself this is complete. I'm told that $$G/\Phi (G)$$ can be of prime powered order. I'm afraid i might be missing something.

 

[mighty2000]

Your attempt is on the right track, but there are a few issues with your reasoning.

Firstly, you cannot conclude that G/P is cyclic just because it has order p. For example, the Klein four-group has order 4 but is not cyclic. However, you can conclude that G/P is abelian since it is isomorphic to a subgroup of the abelian group Z/pZ.

Secondly, your claim that there is only one maximal subgroup is not necessarily true. In a finite group, there can be multiple maximal subgroups. However, it is true that \Phi(G) is the intersection of all maximal subgroups.

Finally, you are correct that G/\Phi(G) can have prime powered order. In fact, it will always have prime powered order since G/\Phi(G) is a quotient group of G, and quotient groups always have order that is a divisor of the original group's order.

To complete the proof, you can use the fact that G/\Phi(G) is a quotient group of G to show that it is elementary abelian. Since G is a p-group, every element of G has order that is a power of p. This means that when you take the quotient by \Phi(G), the orders of the elements in the quotient group will also be powers of p. This, combined with the fact that G/\Phi(G) is abelian, implies that G/\Phi(G) is an elementary abelian p-group.