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

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