# How to prove that two Groups Isomorphic?

We know that when two groups are cyclic and have the same order then these two groups are isomorphic. If we know that one Group is cyclic and the other is not and they have the same order could we say that they are not isomorphic to each other?

• Suppose that such a isomorphism exists, say $f:X\to Y$ ! You have $X$ to be cyclic. Then can you show that the $Y$ is cyclic too! – Riju Dec 11 '17 at 11:29
• For your second question, $\mathbb{Z}_{6}$ and $S_{6}$ are not isomorphic. – akech Dec 11 '17 at 11:30
• yes, I know that I should prove that there is f which is homomorphism and Bijection. But I think we could say that two groups are Isomorphic if they have the same order and they are cyclic. – Dania Dec 11 '17 at 11:35
• So could we use that as a proof that they are not Isomorphic? For example, if I have the Unit group of Z/8 which is not cyclic and the unit group of z/5 which is cyclic they both have the same order 4 so could we say that they are not isomorphic? – Dania Dec 11 '17 at 11:38

Note that if $a$ is the generator of a cyclic group,i.e. $G=\langle a\rangle$, then $\phi(G)=\langle \phi(a)\rangle$, for any homomorphism $\phi$.
• If it is an isomorphism then the second group is same as $\phi(G)$, and hence cyclic. – Abishanka Saha Dec 11 '17 at 11:49
"If we know that one Group is cyclic, say $C_4$, and the other is not, say $C_2\times C_2$, and they have the same order, namely $4$, could we say that they are not isomorphic to each other?" Yes, we could. The property of being cyclic, or abelian (or being nilpotent, solvable etc.),is preserved under isomorphism. The idea of an "isomorphism" is, that all "algebraic" properties are preserved.