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Can someone tell me if this proof is sufficient given the assumption. Let H,K be normal subgroups of G. Assumption: g(H∩K)= (gH)∩(gK)

Proof: g(H∩K) =(gH)∩(gK) =(Hg)∩(Kg) =(H∩K)g

Can someone tell me if I am missing something? I haven't seen this solution before and it seems simple.


marked as duplicate by Théophile, Don Thousand, user10354138, Chinnapparaj R, Robert Chamberlain Nov 12 '18 at 9:25

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