Question

If G is an abelian group of order ${\mathbf{p}}^{\mathbf{t}}\mathbf{m}$with role="math" localid="1655708518756" $(p,m)=1$prove that $\mathbf{G}\left(p\right)$has order ${\mathbf{p}}^{\mathbf{t}}$ .

Short answer

It is proved that, $\mathrm{G}\left(\mathrm{p}\right)$has order ${\mathrm{p}}^{\mathrm{t}}$.

Step-by-step solution

Fundamental Theorem of a Finite Abelian Group and Cauchy Theorem for Abelian Groups

Fundamental Theorem of a Finite Abelian Group

Every finite abelian group G is the direct sum of cyclic groups, each of prime power orders.

Cauchy Theorem

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

Given that G is a abelian group of order ${\mathrm{p}}^{\mathrm{t}}\mathrm{m}$ with $\left(\mathrm{p},\mathrm{m}\right)=1$.

 G(p)has order pt

Let the order of G(p) will be $\left|\mathrm{G}\left(\mathrm{p}\right)\right|={\mathrm{p}}^{\mathrm{k}}$.

Then, by Cauchy Theorem if p divides $\left|\mathrm{G}\left(\mathrm{p}\right)\right|$ then, G(p) contains element of order q.

By the definition, G(p) contains only element with order a power of p.

Here, the order of subgroup divides the order of whole group implies $\mathrm{k}\le \mathrm{t}$.

If k=t then, $\left|\mathrm{G}\left(\mathrm{p}\right)\right|={\mathrm{p}}^{\mathrm{t}}$.

The base case at t=0 is trivially true.

If $\mathrm{t}>0$, $\mathrm{k}<\mathrm{t}$ then $\left|\mathrm{G}/\mathrm{G}\left(\mathrm{p}\right)\right|={\mathrm{p}}^{1-\mathrm{k}}\mathrm{m}$.

Since, $\mathrm{t}-\mathrm{k}>0$ , by Cauchy Theorem there exists an element $\mathrm{a}\in \mathrm{G}/\mathrm{G}\left(\mathrm{p}\right)$of order p that is ${\mathrm{a}}^{\mathrm{P}}\in \mathrm{G}\left(\mathrm{p}\right)$ implies that $\mathrm{a}\in \mathrm{G}\left(\mathrm{p}\right)$.

If $\left[\mathrm{a}\right]=\left[\mathrm{e}\right]$then, the order of $\left[\mathrm{a}\right]$ is 1, which is not equal to p is a contradiction.

Thus, k must be equal to t and $\mathrm{G}\left(\mathrm{p}\right)$has order ${\mathrm{p}}^{\mathrm{t}}$.

Therefore, if G is a 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 ${\mathrm{p}}^{\mathrm{t}}$.