Show that 2 group with order 3 is isomorphic Solution I think that you can prove

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I think that you can prove related results in one go. Consider any group of prime order, lets prove that its cyclic Consider G = {e,a, ..} consider the subgroup generated by a ie (a) = {e,a,a^-1, a^2, a^-2....} the order of this grp > 1 as a!= e but order of a subgrp divides the order of the group hence its order must be p and hence (a) = G hence if |G| is prime then its cyclic, ie generated by a single element Now consider 2 groups of order p G,H both are cyclic: and G = (a) and H = (b) take f(a) = b and we can prove that this is a isomorphism. as f(a^i * a^j) = f(a^(i+j)) = b^(i+j) = b^i . b^j also the kernel is f(a^0=e) = e hence its an isomorphism Hence proved message me if you have any doubts

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