Let $G$ be abelian of order $p_1^{m_1}p_2^{m_2}...p_k^{m_k}$. Show $G\cong P_1\times ...\times P_k$ where $P_i$ are subgroups of order $p_i^{m_i}$

Question

I assume I should prove that $G$ is equal to inner product $P_1P_2\cdots P_k$ and that there would be natural isomorphism. Since $G$ is abelian all its subgroups are normal but I don't know why representation via those subgroups is unique?

Answer 1

As said in the comment, there are other questions ($1$,$2$,$3$) regarding the factorization of $G$ in groups of order equal to a prime power. Basically, since $G$ is abelian, for every integer $n\in\mathbb Z$ the map $$g\mapsto g^n$$ is a group homomorphism $G\rightarrow G$, which implies that the image $G^n=\{g^n:g\in G\}$ is a subgroup of $G$.

Now, if $G=nm$ and $n,m$ are relatively prime, then $G\simeq G^n\times G^m$ with an isomorphism given by the map $\phi:G^n\times G^m\ni(g,h)\mapsto gh$

• First of all, notice that $G^n\cap G^m=(e)$ (where $e$ is the identity in $G$), because every element in $G^n$ has order dividing $m$ (it immediate from the definition of $G^n$ that $g^m=1$, for all $g\in G^n$). Also, for the same reason every element in $G^m$ has order dividing $n$. We conclude that an element $g\in G^n\cap G^m$ has order dividing both $n,m$, which implies that $g=1$.

• Given $g,g'\in G^n$ and $h,h'\in G^m$, if $gh=g'h'$, then $(g')^{-1}g=h'h^{-1}$, with $(g')^{-1}g\in G^n$ and $h'h^{-1}\in G^m$, so $$(g')^{-1}g,h'h^{-1}\in G^n\cap G^m=(e)\rightarrow g=g'\text{ and }h=h'$$ we conclude that the map $\phi:G^n\times G^m\rightarrow G$ is 1-1.

• Finally, since $n,m$ are relatively prime, there are $k,l\in\mathbb Z$ such that $$kn+lm=1$$ so, for every $g\in G$, $$g=g^1=g^{kn+lm}=g^{lm}g^{kn}=\phi(g^{lm},g^{kn})$$and $g^{lm}\in G^n,g^{kn}\in G^m$, so $\phi$ is onto as well

So, we conclude that given an abelian group $G$ with order $|G|=nm$, with $n,m$ relatively prime, then $$G\simeq G^n\times G^m$$ applying induction you can prove that if $|G|=p_1^{n_1}\cdots p_k^{n_k}$, then $G\simeq G^{p_1^{n_1}}\times\cdots\times G^{p_k^{n_k}}$

Now, this completes the proof of the isomorphism. Regarding the "naturality" of this isomorphism, you can prove that as well, but if you mean in categorical terms, you need first to specify between what functors is this isomorphism.