# Abstract Nonsense

## Review of Group Theory: Direct Product of Groups (Pt. II)

Point of post: This post is a continuation of this post.

The interesting thing is that these properties characterize the direct product of groups. More specifically:

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Theorem: Let $G_1,\cdots,G_n,P$ be groups and $\phi_k\in\text{Hom}\left(G_k,P\right)$ for $k\in[n]$ be such that $\phi_k\phi_\ell=\phi_\ell\phi_k$ whenever $k\ne\ell$. Assume that the same universal property: if $\theta_k\in\text{Hom}\left(G_k,G\right)$ for each $k\in[n]$ are such that $\theta_k\theta_\ell=\theta_\ell\theta_k$ whenever $k\ne\ell$ then there exists a unique $\theta\in\text{Hom}\left(P,G\right)$ such that $\theta\circ\phi_k=\theta_k$ for each $k\in[n]$. Then, $\displaystyle P\cong\prod_{j=1}^{n}G_j$.

Proof: For lack of diagrams, we leave this as an exercise to the interested reader. $\blacksquare$

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Another equivalent characterization is

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Theorem: Let $G_1,\cdots,G_n$ be groups. Then,  $\displaystyle G\cong\prod_{j=1}^{n}G_j$ if and only if $G$ contains subgroups $H_j\cong G_j,\text{ }j\in[n]$ such that

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\displaystyle \begin{aligned}&\mathbf{(1)}\quad a_ia_j=a_ja_i\text{ for all }a_i\in A_i\text{ and }a_j\in A_j\text{ and }i\ne j\\ &\mathbf{(2)}\quad \text{Every }g\in G\text{ may be uniquely written as }g=a_1\cdots a_n\text{ with }a_j\in H_j\text{ }j\in[n]\end{aligned}

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Proof: Suppose first that $G$ satisfies $\mathbf{(1)}$ and $\mathbf{(2)}$ and define the guaranteed isomorphisms

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$\phi_k:G_k\to H_k$

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Then, consider

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$\displaystyle \Phi:\prod_{j=1}^{n}G_j\to G:(g_1,\cdots,g_n)\mapsto \phi_1(g_1)\cdots\phi_n(g_n)$

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Evidently this an injection by $\mathbf{(1)}$ and a surjection by $\mathbf{(2)}$ and it’s a homomorphism since each of the $\phi_k(g_k)$ commute with the $\phi_\ell(g_\ell)$ when $\ell\ne k$ (using the same proof as in the second universal property of direct products). It follows that $\Phi$ is an isomorphism.

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Conversely, it’s fairly easy to see that the groups $\tilde{G_1},\cdots,\tilde{G_k}$ satisfy $\mathbf{(1)}$ and $\mathbf{(2)}$ for $\displaystyle \prod_{j=1}^{n}G_j$. $\blacksquare$

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We end this post with a few miscellanea:

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Theorem: Let $G_1,G_2,H_1,H_2$ groups. Then:

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\displaystyle \begin{aligned}&\mathbf{(1)}\quad\textit{If }G_i\cong H_i,\text{ }i=1,2\textit{ then }G_1\times H_1\cong G_2\times H_2\\ &\mathbf{(2)}\quad\textit{If }H_i\unlhd G_i\text{ }i=1,2\textit{ then }H_1\times H_2\unlhd G_1\times G_2\textit{ and }(G_1\times G_2)/(H_1\times H_2)\cong (G_1/H_1)\times (G_2/H_2)\\ &\mathbf{(3)}\quad \mathbb{Z}_m\times\mathbb{Z}_n\cong \mathbb{Z}_{mn} \textit{ if and only if }(m,n)=1\end{aligned}

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Proof:

$\mathbf{(1)}$: This follows immediately since it’s clear that if $\phi:G_1\to G_2$ and $\theta:H_1\to H_2$ are isomorphism then $\gamma:G_1\times H_1\to G_2\times H_2:(g,h)\mapsto (\phi(g),\theta(h))$ is an isomorphism.

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$\mathbf{(2)}$: Define

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$\phi:G_1\times G_2\to (G_1/H_1)\times (G_2/H_2):(g_1,g_2)\mapsto \left(g_1H_1,g_2H_2\right)$

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this is evidently surjective, and a homomorphism since

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\begin{aligned}\phi\left((g_1,g_2)(g'_1,g'_2)\right) &=\phi(g_1g'_1,g_2,g'_2)=\left((g_1g'_1)H_1,(g_2g'_2)H_2\right)\\ &=\left(g_1H_1,g_2H_2\right)\left(g'_1H_1,g'_2H_2\right)\\ &=\phi\left(g_1,g_2\right)\phi\left(g'_1,g'_2\right)\end{aligned}

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Noting though that

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\displaystyle \begin{aligned}\ker\phi &= \left\{(g_1,g_2):\left(gH_1,g_2H_2\right)\right\}\\ &= \left\{(g_1,g_2):g_1=H_1\text{ and }g_2H=H_2\right\}\\ &=\left\{(g_1,g_2):g_1\in H_1\text{ and }g_2\in H_2\right\}\\ &= H_1\times H_2\end{aligned}

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allows us to finish the argument by the First Isomorphism Theorem.

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$\mathbf{(3)}$: Suppose that $(m,n)=d?>1$. Then, we have that $\displaystyle \text{l.c.m}(m,n)=\frac{mn}{(m,n)} yet for every $(a,b)\in\mathbb{Z}_1\times\mathbb{Z}_2$ we have that

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$\left(a,b\right)^{\text{l.c.m}(m,n)}=\left(a^{\text{l.c.m}(m,n)},b^{\text{l.c.m}(m,n)}\right)=(e,e)$

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so that $\mathbb{Z}_m\times\mathbb{Z}_n$ is not generated by any of its elements.

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Conversely, note that evidently $(1,1)^{mn}=(0,0)$ and thus $|(1,1)|\mid mn$. That said, note that since

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$(0,0)=(1,1)^{|(1,1)|}=\left(1^{|(1,1)|},1^{|(1,1)|}\right)$

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we have that $m=|1|\mid |(1,1)|$ and $n=|1|\mid |(1,1)|$ and since $(m,n)=1$ we may conclude that $mn\mid |(1,1)|$ from where the conclusion follows.

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References:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200

January 10, 2011 -