Question

If G is a finite abelian group and p is a prime such that ${\mathbf{p}}^{\mathbf{n}}$ divides $\left|G\right|$then prove that $\mathbf{G}$ has a subgroup of order ${\mathbf{p}}^{\mathbf{n}}$.

Short answer

It is proved that, $\mathrm{G}$ has a subgroup of order${\mathrm{p}}^{\mathrm{n}}$ .

Step-by-step solution

Cauchy Theorem and First Isomorphism Theorem

The Cauchy Theorem states that if $\mathrm{G}$is a finite abelian group and $\mathrm{p}$ is a prime that divides $\left|\mathrm{G}\right|$then, $\mathrm{G}$ contains an element of order $\mathrm{p}$.

The First Isomorphism Theorem states that let $\mathrm{f}:\mathrm{G}\to \mathrm{H}$ be a surjective homomorphism of groups with kernel K. Then, the quotient group $\mathrm{G}/\mathrm{H}$ is isomorphic to H.

 G has a subgroup of order  pn

Let us prove this by induction, for $\mathrm{n}=1$.

Then, by $\mathrm{Cauchy}\mathrm{Theorem}$there exists an element $\mathrm{a}\in \mathrm{G}$ of order p such that is a subgroup of order p in$\mathrm{G}$.

Suppose that $\mathrm{n}>1$and $\mathrm{k}<\mathrm{n}$, $\mathrm{K}$ is an abelian group.

Then, by localid="1655712959545" $\mathbf{Cauchy}\mathbf{}\mathbf{Theorem}{p}^k\left|K\right|$ then, K has a subgroup of order ${\mathrm{p}}^{\mathrm{k}}$.

Since $\mathrm{p}\left|\mathrm{G}\right|$, there is an element $\mathrm{a}\in \mathrm{G}$ of order p.

Thus, there is a group such that .

Consider a map the quotient homomorphism.

Then, by First Isomorphism Theorem is the kernel of $\mathrm{\Phi }$ implies $\left|\mathrm{H}\right|={\mathrm{p}}^{\mathrm{n}}$.

Therefore, if G is an abelian group of order ${\mathrm{p}}^{\mathrm{t}}\mathrm{m}$ with $\left(\mathrm{p},\mathrm{m}\right)=1$ then, $\mathrm{G}\left(\mathrm{p}\right)$ has order ${\mathrm{p}}^{\mathrm{t}}$.