Q:

Inner Product of center Z(G) of a Group

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...

A:

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 $\;n-1\;$ .

