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Let $G$ be a group and $\{N_j\}_{j \in J} $ be a family of proper normal subgroups of $G$ such that $G=\cup_{j \in J} N_j$ and $N_i \cap N_j =\{e\}$ for every $i\ne j \in J$ .

Then how to prove that $G$ is abelian ?

I can show that $ab=ba$ whenever $a\in N_i, b \in N_j$ for some $i\ne j$ . So if I can only show each $N_i$ is abelian, we're done. Unfortunately I'm unable to show that.

Please help

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1 Answer 1

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Let $N$ be one of the normal subgroups and let $C$ be its centraliser in $G$. This is a subgroup of $G$ and you have shown that $C\cup N=G$, but a group cannot be the union of two proper subgroups, so $C=G$ and $N$ is central.

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    $\begingroup$ (I think it's worth noting $C\cup N=G$ because $C$ contains all the other $N$s.) $\endgroup$
    – anon
    Oct 25, 2019 at 2:55
  • $\begingroup$ "a group cannot be the union of two proper subgroups" - very simple proof here $\endgroup$
    – user169852
    Oct 25, 2019 at 5:26

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