If H and K are subgroups of ' order p, show that H=K or H∩K=1My thought would be the Lagrange Theorem. If H is a subgroup of order p of group G, then |H| divides |G|. If |G| isdivided by two subgroups with the same order, then the result is the same. The condition where the subgroups are are thesame makes sense (mostly), but the other condition makesno sense.I honestly have no idea how one would prove this.Edit: Prime group is cyclic. Cyclic group generated by a single element. If the groups aren't the same, (H=K), then the only element they have in common is the identity element (in this case 1)?

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If H and K are subgroups of &#039; order p, show that H=K or H∩K=1My thought would be the Lagrange Theorem. If H is a subgroup of order p of group G, then |H| divides |G|. If |G| isdivided by two subgroups with the same order, then the result is the same. The condition where the subgroups are are thesame makes sense (mostly), but the other condition makesno sense.I honestly have no idea how one would prove this.Edit: Prime group is cyclic. Cyclic group generated by a single element. If the groups aren&#039;t the same, (H=K), then the only element they have in common is the identity element (in this case 1)?

lalilulelo2k3eq · Accepted Answer

Use the primality of p and Lagrange&#039;s theorem to conclude that for all h∈H such that h≠1 we haveH={h0,h1,h2,…,hp−1}.In an entirely analogous way for all k∈K such that k≠1 we haveK={k0,k1,k2,…,kp−1}.If u∈H∩K and u≠1 thenK={u0,u1,u2,…,up−1} and H={u0,u1,u2,…,up−1}

Terry Ray · Accepted Answer

Suppose H≠K. We have that H∩K is a subgroup of both H and K. By Lagrange, either |H∩K|=p or |H∩K|=1, but if the former, then we must have H=H∩K=K, a contradiction;   Answer: Hence |H∩K|=1, which can only happen if H∩K=1.

If H and K are subgroups of ' order p, show that H=K\ \text{or}\ H \cap K=1 My thought wou

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