Question

If p and q are primes with $\mathbf{p}\mathbf{<}\mathbf{q}$and $\mathbf{q}\mathbf{\ne }\mathbf{1}\left(\mathrm{mod}p\right)$and G is a group of order data-custom-editor="chemistry" ${\mathbf{p}}^{\mathbf{2}}\mathbf{q}$, prove that G is abelian.

Short answer

It is proved that, G is abelian.

Step-by-step solution

Step 1:Third Sylow Theorem

The number of Sylow p-subgroups in a finite group G divides $\left|G\right|$ and is of the form for some nonnegative integer k.

By above theorem, the number n_p of Sylow p-subgroups divides q is of the form data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{p}}=\mathrm{kp}+1$.

So that data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{p}}=1$or data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{p}}=\mathrm{q}$, and since data-custom-editor="chemistry" $\mathrm{q}\ne 1$mod p, we can conclude that data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{p}}=1$.

We also have that the number n_q of Sylow q-subgroups divides p² and is of the form data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{q}}=\mathrm{kq}+1$, so that data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{q}}=1$or data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{q}}={\mathrm{p}}^2$(we can rule out data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{q}}=\mathrm{p}$since data-custom-editor="chemistry" $\mathrm{p}<\mathrm{q}$).

If data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{q}}={\mathrm{p}}^2$then data-custom-editor="chemistry" $\mathrm{kq}={\mathrm{p}}^2-1=\left(\mathrm{p}-1\right)\left(\mathrm{p}+1\right)$and so q either divides p-1 or p+1.

Since p<q we can not have q/p-1 and if q/p+1 with data-custom-editor="chemistry" $\mathrm{p}<\mathrm{q}$we must have data-custom-editor="chemistry" $\mathrm{q}=\mathrm{p}+1$which is a contradiction to $q\ne 1$mod p.

We conclude that, data-custom-editor="chemistry" ${\mathrm{n}}_{\mathrm{q}}=1$.

Using Corollary 9.16

This says that let G be a finite group and K a Sylow p-subgroup for some prime p. Then K is normal in G if and only if K is the only Sylow p-subgroup in G.

Therefore, G contains a normal Sylow p-subgroup H of order p², which is abelian byCorollary 9.29. It states that if G is a group of order p² with prime p then G is abelian.

And a normal Sylow subgroup K of order q, which must be cyclic and so particularly abelian.

By Exercise 9.3.13,we have that $\mathrm{G}\cong \mathrm{H}\times \mathrm{K}$and products of abelian groups are abelian.

We conclude that, G is abelian.