Question

Prove that a finite abelian $\mathrm{p}$-group has order a power of $\mathrm{p}$.

Short answer

It is proved that, a finite abelian $\mathrm{p}$ -group has order a power of $\mathrm{p}$ .

Step-by-step solution

Fundamental Theorem

Every finite abelian group $\mathrm{G}$ is the direct sum of cyclic groups, each of prime power orders.

Order of power  

Let $\mathrm{G}$ be a finite abelian $\mathrm{p}$-group that is every element in $\mathrm{G}$ has order .

By Fundamental Theorem of abelian group, $\mathrm{G}$ can be written as direct sum of cyclic groups with prime power as $G={\mathrm{\mathbb{Z}}}_{{p_1}^{a_1}}\oplus {\mathrm{\mathbb{Z}}}_{{p_2}^{a_2}}\oplus ....\oplus {\mathrm{\mathbb{Z}}}_{{p_n}^{a_n}}$ .

If role="math" localid="1657300211593" $\mathrm{p}\ne \mathrm{p}$, take the generator role="math" localid="1657300354466" ${\mathrm{g}}_{p_i}\in {\mathrm{Z}}_{{p_i}^{a_i}}$ then, role="math" localid="1657300306373" $\left(0,0,\dots ,{\mathrm{g}}_{{\mathrm{p}}_{\mathrm{i}}},0,0\right)\in \mathrm{G}$has order ${p_i}^{a_i}$ as $g_{p_i}$ which is a contradiction to the definition of $\mathrm{p}$-group.

Thus, $p_i=p$ for all $\mathrm{I}$ and the order of group $\mathrm{G}$ can be written as:

role="math" localid="1657300948976" $\begin{array}{l}\left|\mathrm{G}\right|=\left|{\mathrm{\mathbb{Z}}}_{{{\mathrm{p}}_1}^{{\mathrm{a}}_1}\oplus }{\mathrm{\mathbb{Z}}}_{{{\mathrm{p}}_2}^{{\mathrm{a}}_2}\oplus }...{\mathrm{\mathbb{Z}}}_{{\mathrm{pn}}^{{\mathrm{a}}_{\mathrm{n}}}}\right|\\ =\left|{\mathrm{\mathbb{Z}}}_{{{\mathrm{p}}_1}^{{\mathrm{a}}_1}}\right|\left|{\mathrm{\mathbb{Z}}}_{{{\mathrm{p}}_2}^{{\mathrm{a}}_2}}\right|....\left|{\mathrm{\mathbb{Z}}}_{{\mathrm{pn}}^{{\mathrm{a}}_{\mathrm{n}}}}\right|\\ ={{\mathrm{p}}_1}^{{\mathrm{a}}_1}{{\mathrm{p}}_2}^{{\mathrm{a}}_2}...{{\mathrm{p}}_{\mathrm{n}}}^{{\mathrm{a}}_{\mathrm{n}}}\end{array}$

Since role="math" localid="1657300955154" ${\mathrm{p}}_{\mathrm{i}}=\mathrm{p}$, the order of $\mathrm{G}$ is role="math" localid="1657301007404" $\left|\mathrm{G}\right|={\mathrm{p}}^{\mathrm{k}}$ where $\mathrm{k}$is positive.

Therefore, a finite abelian $\mathrm{p}$-group has order a power of $\mathrm{p}$.