Alg7.14

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proof Let H and K, each of prime order p, be subgroups of a group G. If H\neq K, prove H\cap K=<e>.

If H and K are of prime order p, every element h\in H and k\in K have order 1 or order p by the corollary to Lagrange's Theorem. However, only the identity can have order 1 and the identity is unique. Thus both H and K have p-1 elements of order p. Since the orders of those elements are p in a group of order p, both H and K have p-1 elements which generate their respective groups. Now, since H\neq K and any k\in K, k\neq e, generates exactly K, k\notin H. Likewise for any h\in H, h\neq e, h\notin K. Since H and K are both groups, they both must contain the identity e. Therefore, if H\neq K, where H and K are of prime order p, it must be true that H\cap K=<e>.


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