Question

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

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 $\mathrm{K}$. Then, the quotient group $\mathrm{G}/\mathrm{K}$ is isomorphic to $\mathrm{H}$.

 G has a subgroup of order pn

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

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

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

Then, by Cauchy Theorem ${\mathrm{p}}^{\mathrm{k}}|\left|\mathrm{K}\right|$ then, $\mathrm{K}$ 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 $\mathrm{p}$.

Thus, there is a group such that .

Consider a map the quotient homomorphism.

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

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