Question

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

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 G is the direct sum of cyclic groups, each of prime power orders.

Order of power p

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

By Fundamental Theorem of abelian group, G can be written as direct sum of cyclic groups with prime power as $\mathrm{G}={\cancel{\mathrm{c}}}_{{{\mathrm{p}}_1}^{{\mathrm{a}}_1}}\oplus {\cancel{\mathrm{c}}}_{{{\mathrm{p}}_2}^{{\mathrm{a}}_2}}\oplus \mathrm{L}\oplus {\cancel{\mathrm{c}}}_{{{\mathrm{p}}_{\mathrm{n}}}^{{\mathrm{a}}_{\mathrm{n}}}}$.

If ${\mathrm{p}}_{\mathrm{i}}=\mathrm{p}$, take the generator ${\mathrm{g}}_{{\mathrm{p}}_{\mathrm{i}}}\in {\cancel{\mathrm{c}}}_{{{\mathrm{p}}_{\mathrm{i}}}^{{\mathrm{a}}_{\mathrm{i}}}}$then, $\left(0,0,\mathrm{K},{\mathrm{g}}_{{\mathrm{p}}_{\mathrm{i}}},0,0\right)\in \mathrm{G}$ has order ${{\mathrm{p}}_{\mathrm{i}}}^{{\mathrm{a}}_{\mathrm{i}}}$ as ${\mathrm{g}}_{{\mathrm{p}}_{\mathrm{i}}}$ which is a contradiction to the definition of p-group.

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

$\begin{array}{rcl}\left|\mathrm{G}\right|& =& \left|{\cancel{\mathrm{c}}}_{{{\mathrm{p}}_1}^{{\mathrm{a}}_1}}\oplus {\cancel{\mathrm{c}}}_{{{\mathrm{p}}_2}^{{\mathrm{a}}_2}}\oplus \mathrm{L}\oplus {\cancel{\mathrm{c}}}_{{{\mathrm{p}}_{\mathrm{n}}}^{{\mathrm{a}}_{\mathrm{n}}}}\right|\\ & =& \left|{\cancel{\mathrm{c}}}_{{{\mathrm{p}}_1}^{{\mathrm{a}}_1}}\right|\left|{\cancel{\mathrm{c}}}_{{{\mathrm{p}}_2}^{{\mathrm{a}}_2}}\right|\mathrm{K}\left|{\cancel{\mathrm{c}}}_{{{\mathrm{p}}_{\mathrm{n}}}^{{\mathrm{a}}_{\mathrm{n}}}}\right|\\ & & {{=\mathrm{p}}_1}^{{\mathrm{a}}_1}{{\mathrm{p}}_2}^{{\mathrm{a}}_2}{\mathrm{K}{\mathrm{p}}_{\mathrm{n}}}^{{\mathrm{a}}_{\mathrm{n}}}\end{array}$

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

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