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We are given that G is a group and a∈G such that |a|=30. Also, H=⟨a⟩ and K=⟨a4⟩. We need to find the number of distinct cosets that K has in H and identify these cosets.The index of K in H is given by [H:K]=|H|/|K|. We can find |H| as |H|=|a|=30. To find |K|, note that ⟨a4⟩={a4k:k∈ℤ}. So, |K|=|a||⟨a4⟩|.Since a4k generates ⟨a4⟩, we have |⟨a4⟩|=|⟨a4k⟩|. Also, a4k has order 30/gcd(4k,30) in G. Since 4 and 30 are relatively prime, we have gcd(4k,30)=gcd(k,15). Thus, a4k has order 30/dk where dk=gcd(k,15).Now, |K|=|a||⟨a4⟩|=30|a4k|=3030/dk=dk.Thus, [H:K]=|H||K|=30dk. We need to find the distinct cosets of K in H. Let x∈H be arbitrary. Then, x=ak for some 0≤k≤29. The coset of K containing x is given by xK={xk:k∈K}.Since K=⟨a4⟩, we have K={1,a4,a8,…,a28}. Thus, xK={ak,ak+4,ak+8,…,ak+24}.Since k can be any integer from 0 to 29, we have 30 distinct cosets of K in H. These cosets are given by:K,aK,a2K,…,a28K.

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