Question

If $\mathbf{G}$is an abelian group of order $\mathbf{n}$ and $\mathbf{k}$, prove that there exist a group $\mathbf{H}$ of order $\mathbf{k}$ and a surjective homomorphism $\mathbf{G}\mathbf{\to }\mathbf{H}$ .

Short answer

It is proved that, there exist a group $\mathrm{H}$ of order $\mathrm{k}$ and a surjective homomorphism $\mathrm{G}\to \mathrm{H}$.

Step-by-step solution

Fundamental Theorem of finite abelian group and Theorem 9.9

Fundamental Theorem of Finite Abelian Group

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

Theorem 9.9

If $\mathrm{n}={{\mathrm{p}}_1}^{{\mathrm{n}}_1}{{\mathrm{p}}_2}^{{\mathrm{n}}_2}{...{\mathrm{p}}_{\mathrm{t}}}^{{\mathrm{n}}_{\mathrm{t}}}$ with role="math" localid="1655718942484" ${\mathrm{p}}_1,....,{\mathrm{p}}_{\mathrm{t}}$ distinct primes then, ${\mathrm{Z}}_{\mathrm{n}}\cong {\mathrm{Z}}_{{{\mathrm{p}}_1}^{{\mathrm{n}}_1}}\oplus ...\oplus {\mathrm{Z}}_{{{\mathrm{p}}_{\mathrm{t}}}^{{\mathrm{n}}_{\mathrm{t}}}}$.

Given that $\mathrm{G}$ is an abelian group of order $\mathrm{n}$ and $\mathrm{k}\overline{)\mathrm{n}}$.

A group H of order k

Let $\mathrm{G}={\oplus }_{\mathrm{i}=1}^{\mathrm{r}}{\mathrm{\mathbb{Z}}}_{{{\mathrm{p}}_{\mathrm{i}}}^{{\mathrm{\alpha }}_{\mathrm{i}}}}$ be an finite abelian group.

Each cyclic group in G generated by ${\mathrm{a}}_{\mathrm{i}}$ and the order of n is role="math" localid="1655719844310" $\mathrm{n}={\mathrm{\pi }}_{\mathrm{i}=1}^{\mathrm{r}}{{\mathrm{p}}_{\mathrm{i}}}^{{\mathrm{\alpha }}_{\mathrm{i}}}$.

Since $\mathrm{k}\overline{)\mathrm{n}}$, $\mathrm{k}={\mathrm{\pi }}_{\mathrm{i}=1}^{\mathrm{r}}{{\mathrm{p}}_{\mathrm{i}}}^{{\mathrm{\beta }}_{\mathrm{i}}}$ with ${\mathrm{\beta }}_{\mathrm{i}}\le {\mathrm{\alpha }}_{\mathrm{i}}$ for each i.

Thus, there exists a subgroup has order $\frac{\mathrm{n}}{\mathrm{k}}={\mathrm{\pi }}_{\mathrm{i}=1}^{\mathrm{r}}{{\mathrm{p}}_{\mathrm{i}}}^{{\mathrm{\alpha }}_{\mathrm{i}}-{\mathrm{\beta }}_{\mathrm{i}}}$ and a surjective homomorphism $\mathrm{\Phi }:\mathrm{G}\to \mathrm{G}/\mathrm{K}$ such that $\left|\mathrm{G}/\mathrm{K}\right|=\mathrm{k}$.

Therefore, if $\mathrm{G}$ is an abelian group of order $\mathrm{n}$ and $\mathrm{k}\overline{)\mathrm{n}}$ then there exist a group $\mathrm{H}$ of order $\mathrm{k}$ and a surjective homomorphism $\mathrm{G}\to \mathrm{H}$.$$