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

I proved $J_n$ is a normal subgroup. I cannot conclude whether my option 4 is correct or not.


It is a subgroup, the question is of what group? And following your notation we look at the direct product of $\;G\;$ with itself $\;n\;$ times:

$$G^{\times n}:=\overbrace{G\times G\times\ldots\times G}^{n\;\text{times}}\implies J_n\lhd \left(Z(G)\right)^{\times n}$$

About the last part: define

$$\phi:J_n\to \left(Z(G)\right)^{\times(n-1)}\;\;,\;\;\phi(g_1,...,g_n):=(g_1,...,g_{n-1})$$

Note the above is well defined since for $\;g_1,...,g_n\in Z(G)\;$ , we have

$$(g_1,...,g_{n-1},g_n)\in J_n\iff g_n=(g_1\cdot...\cdot g_{n-1})^{-1}$$

(Perhaps this is what the other answer tried to convey)

Prove now the above is an isomorphism.


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