Let H and K both be subgroups of G. Show that for all a ∈ G we have Ha ∩ Ka = (H ∩ K)a?
Considering H and K are subgroups I thought there could be two general cases; H ∩ K = {1} or H ∩ K not equal to {1} (i.e. something of higher order)
For H ∩ K = {1}, Clearly (H ∩ K)a = 1a = G. But how do I show Ha ∩ Ka = G??
For the other case; if x ∈ H and x ∈ K then surly x ∈ H ∩ K. So, xa ∈ (H ∩ K)a. Then, xa ∈ Ha and xa ∈ Ka. Thus, xa ∈ Ha ∩ Ka.
I'm not sure if this is even the right way to attack this question; can anybody help?