Let $G$ be a group and $Z(G)$ be its center. For $n\in \mathbb{N}$, define $$J_n=\{(g_1,g_2,...,g_n)\in Z(G)\times Z(G)\times\cdots\times Z(G): g_1g_2\cdots g_n=e\}.$$ As a subset of the direct product group $G \times G \times \dots \times G$ ,

$J_n$ is

(1) not necessarily a subgroup,

(2) a subgroup but not necessarily a normal subgroup,

(3) a normal subgroup,

(4) isomorphic to $Z(G)\times Z(G)\times\cdots\times Z(G)$ $((n-1)$ times).

is $J_n$ a subgroup?

My argument:

let us assume that $n = 2$, then

$J_2 = \{(g_1,g_2)\in Z(G)\times Z(G) :g_1g_2=e\}$ (i.e)

$J_2 = \{(g_1,g_1),(g_1,g_2),(g_2,g_1),(g_2,g_1)\}$ and $g_1g_1= g_1g_2= g_2g_1= g_2g_2=e$

Is this possible? Now consider $Z(G)$

since $g_1$ and $g_2 \in Z(G), g_1g_2=g_2g_1$ and $g_1g_2=g_2g_1 = e$

therefore, $g_1$ is inverse to $g_2$ and vice versa in $Z(G)$, but according to the definition of the set $J_2, g_1g_1=e$ and $g_2g_2=e$ which implies that $g_1$ and $g_2$ are inverses to themselves, but $Z(G)$ is a subgroup every element in it has a unique inverse, hence $J_2$ is not a subgroup....

If there are mistakes in my argument, please tell me where i am wrong...

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    $\begingroup$ In (1)-(3) I assume you mean "a subgroup"...of the cartesian product = direct product $\;G\times G\times\ldots\times G\;$ ? $\endgroup$ – DonAntonio Apr 29 '16 at 8:42
  • $\begingroup$ You can also write $\bigoplus_i^{n} Z(G)$ to make it easier :) Or use prod! $\endgroup$ – Zelos Malum Apr 29 '16 at 8:58
  • $\begingroup$ @Joanpemo Sorry i cant get you.. $\endgroup$ – Sam Christopher Apr 29 '16 at 9:28
  • $\begingroup$ @SamChristopher When you talk of a "subgroup" it is because there is an "overgroup" or larger group in which the assumed subgroup lives, right? $\endgroup$ – DonAntonio Apr 29 '16 at 9:31
  • $\begingroup$ Yes...You are right...sorry..it is actually a subset of GXGX...G(n times)..To prove that J_n is first of all subgroup of GXGX...G then to prove it is normal. $\endgroup$ – Sam Christopher Apr 29 '16 at 9:38

You have to show $\;J_n\;$ isn't empty (trivial), and also

$$(g_1,...,g_n),\,\,(h_1,...,h_n)\in J_n\implies (g_1,...,g_n)(h_1,...,h_n)^{-1}\in J_n$$

But assuming what I asked you in my comment above, this is easy:

$$(h_1,...,h_n)^{-1}=(h_1^{-1},...,h_n^{-1})\implies (g_1,...,g_n)(h_1,...,h_n)^{-1}=$$$${}$$

$$=(g_1h_1^{-1},...,g_nh_n^{-1})\in J_n\iff g_1h_1^{-1}\cdot...\cdot g_nh_n^{-1}=1$$

But each and every one of all these elements commute with everything, and since also

$$h_1^{-1}\cdot\ldots\cdot h_n^{-1}=(h_1\cdot\ldots\cdot h_n)^{-1}=1^{-1}=1$$ we thus get $$g_1h_1^{-1},...,g_nh_n^{-1}=g_1\cdot\ldots\cdot g_n\cdot h_1^{-1}\cdot\ldots\cdot h_n^{-1}=1\cdot1=1$$

So you get (1) and with a very little more work also (3), and thus (2) isn't true, and (4) is true as the last coordinate in $\;J_n\;$ is determined by the first $\;n-1\;$ .

  • $\begingroup$ Thanks for the Answer, I got it..please give me the hint to prove that it is a normal subgroup.. $\endgroup$ – Sam Christopher Apr 29 '16 at 9:35
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    $\begingroup$ Why is (4) clearly false in the finite case? The map $(g_1,\ldots, g_{n-1})\mapsto (g_1,\ldots, g_{n-1},(g_1g_2\cdots g_{n-1})^{-1})$ defines a bijection, which looks to be a homomorphism. $\endgroup$ – Aaron Apr 29 '16 at 9:35
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    $\begingroup$ @SamChristopher Every subgroup of an abelian group is normal. $\endgroup$ – Aaron Apr 29 '16 at 9:38
  • $\begingroup$ So you are saying GXGX..G is abelian...how is that? pls explain $\endgroup$ – Sam Christopher Apr 29 '16 at 9:40
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    $\begingroup$ @SamChristopher The reason it is normal is because for $g \in G, h \in Z(g)$, we have $ghg^{-1}=h$, so if you try to conjugate an element in $J_n$ by an element of $G^n$, you won't change the element (conjugation acts as the identity automorphism, which is a stronger statement than normality). The same proof gives that $Z(G^n)=Z(G)^n$. . $\endgroup$ – Aaron Apr 29 '16 at 10:01

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