Question

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

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

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

Prove H≅K 

By Fundamental Theorem of Finite Abelian Group, $\mathrm{G}$, $\mathrm{H}$ and $\mathrm{K}$ can be written as $G={\oplus }_{i=1}^k{\mathrm{\mathbb{Z}}}_{{p_1}^{a_1}}$, $H={\oplus }_{j=1}^l{\mathrm{\mathbb{Z}}}_{{p_1}^{{\beta }_1}}$ and role="math" localid="1657360029930" $K={\oplus }_{k=1}^m{\mathrm{\mathbb{Z}}}_{{p_k}^{{\gamma }_k}}$.

Here, the set of elementary divisors are $S=\left\{p_i^{{\alpha }_i},1\le i\le k\right\}$, $T=\left\{p_i^{{\beta }_i},1\le j\le 1\right\}$and $U=\left\{p_k^{{\gamma }_k},1\le k\le m\right\}$.

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 $\mathrm{U}\cup \mathrm{U}$.

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

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

Therefore, if $\text{G⊕H≅G⊕K}$ then, $\mathrm{H}\cong \mathrm{K}$.