- #1

Mr Davis 97

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## Homework Statement

Prove that the set of all subgroups of a group ##G## is a lattice with respect to the partial order relation given by containment.

**Note:**You need not prove that containment is a partial order relation but you do need to prove that if ##H\leq G## and ##K\leq G## then there is a unique supremum and a unique infimum in the set of subgrups of ##G## for ##H## and ##K## with respect to containment. In particular, prove that the supremum is ##\langle H\cup K \rangle## and the infimum is ##H\cap K##.

**2. Relevant information**

A supremum for elements ##a,b## of a poset ##(X,\leq)## is an element ##s\in X## such that ##a\leq s, b\leq s## and for any ##s'\in X## satisfying ##a\leq s', b\leq s'## we have ##s\leq s'##.

An infimum for elements ##a,b## of a set ##(X,\leq)## is an element ##i\in X## such that ##i\leq a, i\leq b## and for any ##i'\in X## satisfying ##i'\leq a, i'\leq b## we have ##i'\leq i##.

## The Attempt at a Solution

Here is my attempt:

Let ##H,K## be subgroups of ##G##.

First we show that ##\inf (H,K) = H \cap K##. Clearly ##H \cap K \subseteq H## and ##H \cap K \subseteq K##. Now, by set theory, ##H \cap K## is the largest subset contained in both ##H## and ##K## in the sense that if ##S## is a lower bound for ##\{H,K\}## (i.e. ##S \subseteq H## and ##S \subseteq K##), then ##S \subseteq H \cap K##. This satisfies the definition of infimum.

Second, we want to show that ##\sup(H,K) = \langle H \cup K \rangle##. Clearly ##H,K \subseteq \langle H \cup K \rangle##, so##\langle H \cup K \rangle## is an upper bound of ##\{H,K\}##. Now, let ##S## be an arbitrary upper bound for ##\{H,K\}##. We want to show that ##\langle H \cup K \rangle \subseteq S##...... This is where I get stuck. Any ideas?