Let G = Q8 (quaternion group). Find the right cosets of H in G for H = < J > and H = < -I >

Question

For H = < J > , H = {J, -J, I, -I}, and H*K = {L, -K, -L, K}

For H = < -I >, would H be all of Q8?

Answer 1

Of course not. The quaternion group is not cyclic, it is even not abelian. Actually $-I$ is an element of order $2$ and hence $H=\{I,-I\}$.