Question

Let $\mathbf{G}\mathbf{,}\mathbf{H}\mathbf{,}\mathbf{K}$be finite abelian groups.

If $\mathbf{G}\mathbf{\oplus }\mathbf{G}\mathbf{\cong }\mathbf{H}\mathbf{\oplus }\mathbf{H}$, prove that $\mathbf{G}\mathbf{\cong }\mathbf{H}$ .

Short answer

It is proved that, $\mathrm{G}\cong \mathrm{H}$.

Step-by-step solution

 Step 1: The Fundamental Theorem of Finite Abelian Group and Theorem 9.12

Fundamental Theorem of Finite Abelian Group

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

Theorem 9.12

If G and H are finite abelian groups then, G is isomorphic to H if and only if G and H have the same elementary divisors.

Given that $\mathrm{G}\oplus \mathrm{G}\cong \mathrm{H}\oplus \mathrm{H}$ .

Prove G≅H

By Fundamental Theorem of Finite Abelian Group, the group G and H can be written as $\mathrm{G}={\oplus }_{\mathrm{i}=1}^{\mathrm{k}}{\cancel{\mathrm{c}}}_{{{\mathrm{p}}_{\mathrm{i}}}^{{\mathrm{\alpha }}_{\mathrm{i}}}}$, $\mathrm{H}={\oplus }_{\mathrm{j}=1}^{\mathrm{l}}{\cancel{\mathrm{c}}}_{{{\mathrm{p}}_{\mathrm{j}}}^{{\mathrm{\beta }}_{\mathrm{j}}}}$.

Here, the set of elementary divisors are $\mathrm{S}=\left\{{{\mathrm{p}}_{\mathrm{i}}}^{{}_{\mathrm{\alpha i}}};1\le \mathrm{i}\le \mathrm{k}\right\}$ and $\mathrm{T}=\left\{{{\mathrm{p}}_{\mathrm{i}}}^{{\mathrm{\beta }}_{\mathrm{i}}};1\le \mathrm{j}\le \mathrm{l}\right\}$.

The set of elementary divisors of $\mathrm{G}\oplus \mathrm{G}$ is the disjoint union $\mathrm{S}\cup \mathrm{S}$ and similarly, the set of elementary divisors of $\mathrm{H}\oplus \mathrm{H}$ is the disjoint union $\mathrm{T}\cup \mathrm{T}$.

Thus, by Theorem 9.12, it can be written as $\mathrm{S}\cup \mathrm{S}=\mathrm{T}\cup \mathrm{T}$, which implies $\mathrm{S}=\mathrm{T}$.

Again by Theorem 9.12, $\mathrm{S}=\mathrm{T}$ implies $\mathrm{G}\cong \mathrm{H}$.

Therefore, if $\mathrm{G}\oplus \mathrm{G}\cong \mathrm{H}\oplus \mathrm{H}$ then, $\mathrm{G}\cong \mathrm{H}$.