The Frattini subgroup of $\Bbb{Z}_p \times\Bbb Z _{p^2}.$

Question

Can anyone please help me to find the Frattini subgroup of $\mathbb{Z}_p \times \Bbb Z _{p^2}$? I know that as a set the Frattini subgroup is the set of all non-generators. Is this the only way to compute such subgroups? Is there any better way? My professor's answer was negative.

I believe there should be some cool way to tackle this problem. Can we solve this problem from the definition? I mean by listing all the maximal subgroups, and taking their intersection?

Thanks so much.

Answer 1

The Frattini subgroup of $\mathbf{Z}_p \times \mathbf{Z}_{p^2}$ is equivalently its Jacobson radical as a ring (this is actually true for any ring that is a finite direct product of quotient rings of $\mathbf{Z}$). Since it is a finite ring, the Jacobson radical is the same as the nilradical, which is $\{0\}\times p\mathbf{Z}_{p^2}$. Hence, the Frattini subgroup of $\mathbf{Z}_p \times \mathbf{Z}_{p^2}$ is also $\{0\}\times p\mathbf{Z}_{p^2}$.

Answer 2

One way to solve this problem is to know that the Frattini subgroup behaves well with respect to direct products. That is, for finite groups $A$ and $B$ we have $\Phi(A \times B) = \Phi(A) \times \Phi(B)$. Applying this with $A=C_p$ and $B=C_{p^2}$ yields $\Phi(C_p \times C_{p^2}) = K$, where $K$ is the unique subgroup of order $p$ of $B$.

Working with the definition: note that $\Phi(C_p \times C_{p^2})>1$ and that $C_p \times K$ and $1 \times C_{p^2}$ are both maximal subgroups, thus $\Phi(C_p \times C_{p^2}) \leq K = \left(C_p \times K\right) \cap \left(1 \times C_{p^2}\right)$. It follows that $\Phi(C_p \times C_{p^2}) = K$.