Abstract Nonsense

Crushing one theorem at a time

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)}<mn 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


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January 10, 2011 - Posted by | Algebra, Group Theory | , ,

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