(January 2012) Let \(G\) be a group, and let \(H,K\subseteq G\) be subgroups of \(G\). Prove the following standard theorem about subgroups: that \(H\cap K\) is a subgroup of \(G\).
Solution
Proof. (Nonempty) Since H and K are subgroups of G, we have e∈H and e∈K. So e∈H∩K, and hence H∩K⁄= ∅. (Closed under∗) Given a,b∈H∩K, we have ab∈H and ab∈K since H and K are closed under the operation. So ab∈H∩K, as desired. (Closed under−1) Givena∈H∩K, we havea−1∈H anda−1∈K sinceH andK are closed under inverses. So ab∈H∩K as desired. Thus,H∩K is a subgroup of G. QED See 3 for another problem where you have to prove that a subset is a subgroup. Cyclic Subroups. An element g∈G generates the cyclic subgroup ⟨g⟩ ={gm|m∈ Z}. There are many therorems about cyclic groups, but for the algebra exam, one key fact to know is that g has finite order if and only if⟨g⟩ is finite, in which case o(g) =o(⟨g⟩). [Do you know why this fact is true?] See 2 and 6 for problems involving cyclic subgroups. Lagrange’s Theorem. This theorem says that if H is a subgroup of a finite group G, then the order of H divides the order of G. That is, |H| ⏐⏐|G|. When g is an element of a finite group G, Lagrange’s Theorem has several important consequences: • If|G| is prime, then G is cyclic. • For allg∈G, we have o(g) ⏐⏐|G|. • For allg∈G, we have g|G| =e. These consequences all follow from Lagrange, using the fact that for any g∈ G, we have o(g) =|⟨g⟩|. [Be sure you know how to use this fact to prove the statements above!] In fact, on the comps, if you have a finite group G and want to conclude any of the three bulleted statements above, you may justify simply by saying “by Lagrange’s Theorem.” See 2 , 8 , and 9 for problems that use these corollary versions of Lagrange’s Theorem.