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## Homework Statement

"Write out all the right cosets of H in G where G = (a) is a cyclic group of order 10 and H = (a^2) is the subgroup of G generated by a^2."

## Homework Equations

- If G = (a), then G = {a^i | i=0,-1,1,-2,2...}.

- A right coset is the set Hb = {hb | h is in H}

- Order of G is 10, so G has ten elements. Order of H has to divide that.

- G is the disjoint union of all its right cosets.

- The right cosets of H in G are disjoint.

- The

## The Attempt at a Solution

I guess I'm a bit confused, due to the order being ten but not being given any elements to work with.

I reasoned that H = {a^(2i) | i=0,-1,1...}, which led to me thinking that at least one right coset was:

Ha = {a^(3i) | i = 0, -1, 1,...}

But I'm not sure where I should stop with the i's, leading to me being confused as to how many elements would be in that right coset (if that even is one of the right cosets).

Am I even on the right track?