Why is $Z_2 \times Z_4$ not isomorphic to $Z_8$?

$Z_2 \times Z_4$ has 8 elements and obviously so does $Z_8$. How do you conclude they are not isomorphic then? They essentially have a $1-1$ relationship

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Does $Z_2\times Z_4$ have an element of order $8$? – jwsiegel May 3 '14 at 19:52
– Belgi May 3 '14 at 19:54
They do of course have a 1-1 relationship. However, that relationship is not a group homomorphism. – fkraiem May 3 '14 at 19:55

Indeed, they do have a one-to-one correspondence between, them (in fact, there are $8!$ such correspondences). The problem is that they don't have the "same structure," so none of these correspondences will be homomorphisms, and so no isomorphism exists. But it's tedious in the extreme to check so many functions to see if any of them is a homomorphism, so we instead make use of one of many properties preserved under isomorphisms. Two such (related) properties are that:

1. isomorphisms preserve orders of elements (meaning that any given element will be mapped to an element of the same order), and
2. isomorphic images of cyclic groups are again cyclic groups.

Those two properties are probably the simplest way to show that these two groups are not isomorphic, since (in particular) there are four different order $8$ elements of $\Bbb Z_8,$ but no such elements in $\Bbb Z_2\times\Bbb Z_4.$ So, we can directly conclude that they are not isomorphic by fact 1. Alternately, we can note that this means $\Bbb Z_8$ is cyclic and $\Bbb Z_2\times\Bbb Z_4$ is not, so can indirectly conclude that they are not isomorphic by fact 2.

There are many other such properties that we can use in other cases. It's good to keep several of them in mind. As Mark points out in the comments of the first answer, though, the first property is a necessary condition for a function to be an isomorphism, but not sufficient. However, two cyclic groups of the same order will necessarily be isomorphic. Unfortunately, most groups aren't cyclic, so this property won't usually be useful.

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$\mathbb{Z}_8$ has an element of order $8$, while $\mathbb{Z}_2\times\mathbb{Z}_4$ does not.

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Ah, so all the orders present in one group, must also be there in the second group? – atherton May 3 '14 at 19:54
@RoyM. - Yes, if they are isomorphic – Belgi May 3 '14 at 19:55
@RoyM. If two groups are ismorphic, then they are structurally equivelant. – ellya May 3 '14 at 20:02
Note that it is necessary that the groups have the same number of elements of each order. However that is not a sufficient condition for groups to be isomorphic. See for example math.stackexchange.com/questions/62331/… – Mark Bennet May 3 '14 at 20:04

The range of every homomorphism $f:\mathbb{Z_2 \times Z_4 \rightarrow Z_8}$ is {0,2,4,6} or {0,4}.

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Should have added more explaination, I'd say. – BigbearZzz Apr 13 at 8:24