Question

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

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

Short answer

It is proved that, $G\cong H$.

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 $\mathrm{G}$ is the direct sum of the cyclic group each of prime power.

Theorem 9.12

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

Given that $G\oplus G\cong H\oplus H$

Prove G≅H 

By Fundamental Theorem of Finite Abelian Group, the group $\mathrm{G}$ and $\mathrm{H}$ can be written as $G={\oplus }_{i=1}^k{Z}_{{p_1}^{a_1}}$, $G={\oplus }_{j=1}^i{Z}_{{p_1}^{a_1}}$.

Here, the set of elementary divisors are role="math" localid="1657357250536" $\mathrm{S}=\left\{{\mathrm{p}}_{\mathrm{i}}^{{\mathrm{\alpha }}_{\mathrm{i}}},1\le \mathrm{i}\le \mathrm{k}\right\}$and $\mathrm{T}=\{{\mathrm{p}}_{\mathrm{i}}^{{\mathrm{\beta }}_{\mathrm{i}}},1\le \mathrm{j}\le 1]$.

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, $\mathrm{G}\cong \mathrm{H}$.