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Let $Z(G)$ denote the center of a group $G$, let $J_n=Z(G)\times\dots \times Z(G)$, is it true that as a subset of external direct product $G\times\dots\times G$, $J_n$ is a subgroup?normal sybgroup?is it isomorphic to $Z(G)\times\dots \times Z(G)$ ($(n-1))$ times ? I know $Z(G)$ is a subgroup, so I hope $J_n$ will be so, but I am not sure about the other options.thank you for your help. Well, about the isomorphism I think projection map will work?

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1 Answer 1

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Let $G$ and $H$ be groups, let $A<G$ and $B<H$. The group operation in $G\times H$ is defined to be $$(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2).$$ Clearly, the identity of the operation is $(e_G,e_H)$ where $e_G$ is the identity of $G$ and $e_H$ is the identity of $H$, and for any $(g,h)\in G\times H$, its inverse is $(g^{-1},h^{-1})$.

  • Because $A$ is a subgroup of $G$ and $B$ is a subgroup of $H$, we have that $e_G\in A$ and $e_H\in B$, so that $(e_G,e_H)\in A\times B$.
  • For any $(a_1,b_1),(a_2,b_2)\in A\times B$, we have that $a_1,a_2\in A$ and $b_1,b_2\in B$, so that $a_1a_2\in A$ and $b_1b_2\in B$ because $A$ and $B$ are subgroups of $G$ and $H$ respectively, so that $(a_1,b_1)(a_2,b_2)=(a_1a_2,b_1b_2)\in A\times B$.
  • For any $(a,b)\in A\times B$, we have that $a\in A$ and $b\in B$, so that $a^{-1}\in A$ and $b^{-1}\in B$ because $A$ and $B$ are subgroups of $G$ and $H$ respectively, so that $(a^{-1},b^{-1})\in A\times B$.

Thus, $A\times B$ is a subgroup of $G\times H$ for any subgroups $A<G$ and $B< H$. In fact, for any family of groups and subgroups (not just finite ones), the direct product of the subgroups is a subgroup of the direct product of the groups - just modify the above argument.


Now, we consider the case of normal subgroups. Let $N\triangleleft G$, $K\triangleleft H$, so that for any $g\in G$ and $h\in H$, we have $gNg^{-1}=N$ and $hKh^{-1}=K$. For any $(g,h)\in G\times H$, we have that $$(g,h)(N\times K)(g,h)^{-1}=(g,h)(N\times K)(g^{-1},h^{-1})=\{(g,h)(a,b)(g^{-1},h^{-1})\mid (a,b)\in N\times K\}=\{(gag^{-1},hbh^{-1})\mid (a,b)\in N\times K\}\subseteq N\times K$$ and by symmetry we get $(g,h)(N\times K)(g,h)^{-1}=N\times K$. Thus $(N\times K)\triangleleft (G\times H)$.

In fact, for any family of groups and normal subgroups (not just finite ones), the direct product of the normal subgroups is a normal subgroup of the direct product of the groups - just modify the above argument. Or, you can use induction to show that this is true for a finite collection of groups $G_1,\ldots,G_n$ and normal subgroups $N_1\triangleleft G_1,\ldots,N_n\triangleleft G_n$. Letting $G_1=G_2=\cdots=G_n=G$ and $N_1=N_2=\cdots=N_n=Z(G)$, you have that $J_n$ is a normal subgroup of $G\times\cdots\times G$.


It certainly need not be the case that $J_n=\underbrace{Z(G)\times\cdots\times Z(G)}_{n\text{ times}}$ is isomorphic to $\underbrace{Z(G)\times\cdots\times Z(G)}_{n-1\text{ times}}$. They don't even necessarily have the same cardinality. For example, letting $G=C_2$ be the cyclic group of order 2, so that $|G|=|Z(G)|=2$, then the former has $2^n$ elements and the latter has $2^{n-1}$ elements.

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thank you very much for the answer in so nicely. –  Bunuelian Trick Jun 13 '12 at 9:20
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@Zev, don't you want to add that $Z(G \times \cdot\cdot\cdot \times G) = Z(G) \times \cdot\cdot\cdot \times Z(G)$? –  Nicky Hekster Jun 13 '12 at 19:44

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