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Let $G$ be a group, and $C$ a set of proper subgroups of $G$.

Each subgroup in $C$ is normal subgroup of $G$.

For $G_1 , G_2\in C$, if $G_1 \ne G_2$ then $G_1\cap G_2=\{e_G\}$

$\bigcup\limits_{H\in C}H= G$.

Need to prove that G is Abelian group, hint someone?

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    It's not clear to me what you mean: is C a *set* of **normal** subgroups of G? Or **the set** of all normal subgroups of G? I'm not even sure there's a difference... – 2017-02-20
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    it is some set that apply the condition, it is not have to be the set of all normal sub group of G – 2017-02-20
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    “For $G_1,G_2\in C$, if $G_1\ne G_2$, then $G_1\cap G_2=\{e_G\}$”. Otherwise, the set $C$ would only contain $\{e_G\}$. – 2017-02-20

2 Answers 2

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First note that elements from two different normal subgroups in the family $\mathcal{C}$ commute. If $a \in A$, $b \in B$, with $A, B$ different normal subgroups in $\mathcal{C}$, we have $$ [a, b] = a^{-1} b^{-1} a b = a^{-1} a^{b} = (b^{-1})^{a} b \in A \cap B = 1. $$

Now let $x, y \in A$, with $A \in \mathcal{C}$. Since the subgroups in $\mathcal{C}$ are proper, there is $z \notin A$. So $z$ is contained in a subgroup of the family $\mathcal{C}$ other than $A$, and thus by the above $[x, z] = 1$. Clearly $z y \notin A$, and then again $$ 1 = [x, z y] = [x, y] [x, z]^{y} = [x, y]. $$

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    why (b^(-1))*a*b=a^(x)? – 2017-02-20
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    It's just a notation for the so-called [conjugate](https://en.wikipedia.org/wiki/Conjugacy_class) $a^{b} = b^{-1} a b$. The important fact is that the subgroup $A$ of the group $G$ is a normal subgroup iff for $a \in A$ and $b \in G$ one has $a^{b} \in A$. – 2017-02-20
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Hint: Let $x,y\in G$. Then there exist $G_1,G_2\in C$ such that $x\in G_1,y\in G_2$. Then show $xyx^{-1}y^{-1}\in G_1\cap G_2$

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    The trouble is, when $x, y$ are in the same member of the family. I have addressed this in my answer. – 2017-02-20
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    @AndreasCaranti: You're right! then we have to consider another element outside and manage things. – 2017-02-20