Question

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

If role="math" localid="1656573876136" $\mathbf{G}\mathbf{\oplus }\mathbf{H}\mathbf{\cong }\mathbf{G}\mathbf{\oplus }\mathbf{K}$, prove that $\mathbf{H}\mathbf{\cong }\mathbf{K}$ .

Short answer

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

Step-by-step solution

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 $\mathrm{G}\oplus \mathrm{H}\cong \mathrm{G}\oplus \mathrm{K}$.

Prove  

By Fundamental Theorem of Finite Abelian Group, G, H and K can be written as $\mathrm{G}={\oplus }_{\mathrm{i}=1}^{\mathrm{k}}{\cancel{\mathrm{c}}}_{{{\mathrm{p}}_{\mathrm{i}}}^{{\mathrm{\alpha }}_{\mathrm{i}}}}$, localid="1656574309826" $\mathrm{H}={\oplus }_{\mathrm{j}=1}^{\mathrm{l}}{\cancel{\mathrm{c}}}_{{{\mathrm{p}}_{\mathrm{j}}}^{{\mathrm{\beta }}_{\mathrm{j}}}}$and $\mathrm{K}={\oplus }_{\mathrm{k}=1}^{\mathrm{m}}{\cancel{\mathrm{c}}}_{{{\mathrm{p}}_{\mathrm{k}}}^{{\mathrm{\gamma }}_{\mathrm{k}}}}$.

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

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 $\mathrm{H}\oplus \mathrm{H}$and $\mathrm{K}\oplus \mathrm{K}$ is the disjoint union $\mathrm{T}\cup \mathrm{T}$ and $U\cup U$.

Thus, by Theorem 9.12, $\mathrm{G}\oplus \mathrm{H}\cong \mathrm{G}\oplus \mathrm{K}$can be written as $S\cup T=S\cup U$, which implies $\mathrm{T}=\mathrm{U}$.

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

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