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Is $\mathbb{Z_4}\times\mathbb{Z_6}/\langle(0,1)\rangle$ isomorphic to $\mathbb{Z_6}$? I have counted the order of the former is 4 whereas the order of latter is 6 . Is this the reason that I can conclude they are not isomorphic?

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    $\begingroup$ Of course that's an excellent reason: they don't have the same cardinal, which is a rather trivial consequence (i.e., a necessary condition) of being isomorphic...whic is, among other things, a $\;1-1\;$ map ! $\endgroup$ – DonAntonio Jun 22 at 9:39
  • $\begingroup$ So his two is isomorphic right ? Then how about $\mathbb{Z_6}\times\mathbb{Z_4}/\langle(0,1)\rangle$, are they isomorphic? $\endgroup$ – Ling Min Hao Jun 22 at 9:40
  • $\begingroup$ Deleted my answer because @DonAntonio's is more than sufficient. $\endgroup$ – Ruben Jun 22 at 9:41
  • $\begingroup$ See also this duplicate. $\endgroup$ – Dietrich Burde Jun 22 at 11:30
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If $\;A\cong B\;$ , then there exists a bijective function $\;f:A\to B\implies |A|=|B|\;$ (cardinal = number of elements, if the sets are finite) are the same

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