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Showing that subgroups of G form a lattice

  • #1
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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?
 

Answers and Replies

  • #2
andrewkirk
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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?
 
  • #3
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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##?
 
  • #4
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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?
 
  • #5
andrewkirk
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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.
 

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