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?
Asked
Active
Viewed 38 times
0
-
3Of 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 ! – DonAntonio Jun 22 '19 at 09:39
-
So his two is isomorphic right ? Then how about $\mathbb{Z_6}\times\mathbb{Z_4}/\langle(0,1)\rangle$, are they isomorphic? – Nothing Jun 22 '19 at 09:40
-
Deleted my answer because @DonAntonio's is more than sufficient. – Ruben Jun 22 '19 at 09:41
-
See also this duplicate. – Dietrich Burde Jun 22 '19 at 11:30
1 Answers
1
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
DonAntonio
- 214,715