Currently in Abstract Algebra, discussing group theory. In order to show two groups are isomorphic to each other, I know what you need to show, $1$-$1$, onto, and homomorphism. what I'm having a difficult time doing however is creating a map between the two groups.

Could anyone make any suggestions?

Specifically showing:

$$(Z_2 \times Z_4)/\langle(0,2)\rangle\cong Z_4$$

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    $\begingroup$ Gonna have trouble proving that isomorphism, because they are not isomorphic. Every element $g$ of the left hand group satisfies $g+g=0$. $\endgroup$ – Thomas Andrews May 12 '15 at 17:42
  • $\begingroup$ @ThomasAndrews for sure. Well besides the order or elements in the group what else do I need to be on the look out for? $\endgroup$ – dean3794 May 12 '15 at 17:44
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    $\begingroup$ @San3794 if your group is finite and abelian, then finding the order of elements will always be enough $\endgroup$ – Omnomnomnom May 12 '15 at 17:47
  • $\begingroup$ Order of elements is a big one. Any abelian group can be easily 'factored' as $\mathbb Z_{n_1}\times\mathbb Z_{n_2}\times\cdots\mathbb Z_{n_k}$ with $n_1\mid n_2\mid\cdots\mid n_k$. In the left group, it is $\mathbb Z_2\times\mathbb Z_2$ and the right $\mathbb Z_4$. Commutativity is another - how many pairs $a,b$ such that $ab=ba$. $\endgroup$ – Thomas Andrews May 12 '15 at 17:47
  • $\begingroup$ @ThomasAndrews okay I think I got it... (Z2 x Z4)/ (Z2 x {0}) would be isomorphic to Z4 sry..dont know latex yet! $\endgroup$ – dean3794 May 12 '15 at 18:00

Since those two groups are not isomorphic, you will have trouble finding an isomorphism. For any element $g\in(\mathbb Z_2\times\mathbb Z_4)/\langle(0,2)\rangle$, you have $g+g=0$, while this is not true for $\mathbb Z_4$.

The question of abelian groups turns out to be entirely determined by the number of elements of order $d$ for each $d$.

The general question of determining if two groups are isomorphic (not even finding the isomorphism) is of unknown complexity. As of $2011$, the question has upper bound of $n^{\log n + O(1)}$ time.


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