# Showing that subgroups of G form a lattice

## 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?

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andrewkirk
Homework Helper
Gold Member
You need to show that an arbitrary element of $\langle H\cup K\rangle$ is in S. Since S is an upper bound, H and K are both subgroups of S, and hence subsets of S, so the union is a subset too. An arbitrary element of $\langle H\cup K\rangle$ will be the product of some finite number m of elements of the union. If m =1 is the result easy to prove? What if m=2? Can we extend the result to arbitrary $m\in\mathbb N$ by induction?

You need to show that an arbitrary element of $\langle H\cup K\rangle$ is in S. Since S is an upper bound, H and K are both subgroups of S, and hence subsets of S, so the union is a subset too. An arbitrary element of $\langle H\cup K\rangle$ will be the product of some finite number m of elements of the union. If m =1 is the result easy to prove? What if m=2? Can we extend the result to arbitrary $m\in\mathbb N$ by induction?
Let $g \in \langle H\cup K\rangle$. Then if m = 1, $g$ is either equal to some element in H or some element in K. But both H and K are subsets of S, so then $g \in S$. If m = 2, I feel the only case that's interesting is when we have an element in $H \setminus K$ multiplied by an element in $K \setminus H$... But this product is not guaranteed to be in $H,K$ or $H \cup K$, so how can we show that it must be in $S$?

You need to show that an arbitrary element of $\langle H\cup K\rangle$ is in S. Since S is an upper bound, H and K are both subgroups of S, and hence subsets of S, so the union is a subset too. An arbitrary element of $\langle H\cup K\rangle$ will be the product of some finite number m of elements of the union. If m =1 is the result easy to prove? What if m=2? Can we extend the result to arbitrary $m\in\mathbb N$ by induction?
Actually, I might not even need to prove this... It was remarked in class that $\langle X \rangle$ is the smallest subgroup that contains $X$.

For my proof, would this immediately imply that $\langle H \cup K \rangle \subseteq S$ and so I'm done?

andrewkirk
Let $g \in \langle H\cup K\rangle$. Then if m = 1, $g$ is either equal to some element in H or some element in K. But both H and K are subsets of S, so then $g \in S$. If m = 2, I feel the only case that's interesting is when we have an element in $H \setminus K$ multiplied by an element in $K \setminus H$... But this product is not guaranteed to be in $H,K$ or $H \cup K$, so how can we show that it must be in $S$?
Let the elements be $h\in H$ and $k\in K$. Since $H,K\subseteq S$ we have $h,k\in S$ so $hk\in S$ by group closure axiom applied to S.