## Partitions of a group and normal subgroups

Posted: October 9, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: ,

Let $G$ be a group, $N$ a normal subgroup of $G$ and $P=\{g_iN: \ i \in I \}$ the set of cosets of $N.$ Then $P$ is a partition of $G$ and $g_iN g_jN = g_ig_j N \in P$ for all $i,j \in I.$ We’d like to consider the converse of this.

Problem. Let $G$ be a group and suppose that $P$ is a set partition of $G$ which satisfies the following condition:

$(*)$ for every $Q_1,Q_2 \in P,$ there exists $Q \in P$ such that $Q_1Q_2 \subseteq Q.$

Let $N$ be the element of $P$ which contains the identity element of $G.$ Prove that $N$ is a normal subgroup of $G$ and $P$ is the set of cosets of $N$ in $G.$

Solution. Notice that an obvious result of $(*)$ is that if $Q_1 \in P$ and $g, h \in G,$ then $gQh \subseteq Q_2$ for some $Q_2 \in P.$ Now, if $a \in N,$ then $a = a \cdot 1 \in N^2$ and so $N \subseteq N^2.$ We also have $N^2 \subseteq Q$ for some $Q \in P,$ by $(*).$ Thus $N \subseteq Q$ and so $N=Q.$ Therefore $N=N ^2,$ i.e. $N$ is multiplicatively closed. Let $a \in N.$ Then $Na^{-1} \subseteq Q$ for some $Q \in P$ and since $1 \in Na^{-1},$ we get $1 \in N \cap Q.$ Thus $Q=N$ and so $Na^{-1} \subseteq N.$ Hence $a^{-1} \in N$ and so $N$ is a subgroup. To prove that $N$ is normal, let $g \in G.$ Then, by $(*),$ there exists $Q \in P$ such that $gNg^{-1} \subseteq Q.$ But then $1 \in Q \cap N,$ because $1 \in gNg^{-1} \subseteq Q,$ and so $Q=N.$ Thus $gNg^{-1} \subseteq N,$ i.e.  $N$ is normal. Finally, let $g \in G$ and choose $Q, Q_1 \in P$ such that $g \in Q$ and $gN \subseteq Q_1.$ Then $g \in Q_1 \cap Q,$ because $g \in Ng \subseteq Q_1,$ and so $Q_1=Q.$ Hence $g N \subseteq Q.$ Let $Q_2 \in P$ be such that $N \subseteq g^{-1}Q \subseteq Q_2.$ Then $1 \in Q_2 \cap N,$ because $1 \in g^{-1}Q \subseteq Q_2,$ and so $Q_2=N.$ Hence $gN=Q. \ \Box$