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I have this group of permutations:

permutations

permutes

And I have this group of complex numbers:

complex

These groups are isomorphic to each other, but it seems I do not understand why. I was looking for similarities in structure and I would have thought because they do not share the same structure, that they were not isomorphic. Like take the diagonal of $g$s in the group of permutations. In the group of complex numbers this same diagonal is made of different elements, $i$ and $1$. It seems

I do not fully understand isomorphism, because I was looking for the same structure. Could someone explain why these groups are isomorphic?

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You can see it by constructing an explicit bijective homomorphism.

$e\leftrightarrow 1$, $g\leftrightarrow -1$, and then $f\leftrightarrow i$ or $-i$ and $h$ is whatever remaining that isn't $f$.


Alternatively, you can show that both groups are cyclic of order four.

$i\mapsto i^2=-1\mapsto i^3=-i\mapsto i^4=1\mapsto i^5=i$

$f\mapsto f^2=g\mapsto f^3=h\mapsto f^4=e\mapsto f^5=f$

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  • $\begingroup$ How did you construct the bijection though? $\endgroup$ – Tiernan Watson Jun 13 '16 at 1:50
  • $\begingroup$ @TiernanWatson intuition / observation. I recognized that the tables were exactly alike after some relabeling. More specifically, I noticed that $e$ plays the role of the identity. Next I noticed that $g^2=e$ which acts like $-1$ does in the other table. The last choice could be either. $\endgroup$ – JMoravitz Jun 13 '16 at 1:52
  • $\begingroup$ I can see that if you switch $i$ and $-1$ around you get the same structure that I was initially looking for. I guess it is the uncertainty of accidentally missing it that I do not like because there is no straight method then like a formula to find a bijection. I find working with permutations quite slow as well, so it would have taken me a bit more time to find the generator. $\endgroup$ – Tiernan Watson Jun 13 '16 at 2:09
  • $\begingroup$ The identity should be easy to spot. @TiernanWatson with experience and background knowledge, you will learn that there are only a certain number of possible structures for groups of a specific size. Specifically, for groups of order four, there are only two possible. Either isomorphic to $\Bbb Z_4$ or isomorphic to $\Bbb Z_2\times \Bbb Z_2$ (the klein 4-group). It is clear neither are the latter because the diagonal is not solely the identity. $\endgroup$ – JMoravitz Jun 13 '16 at 2:13
  • $\begingroup$ Since neither can be the klein 4-group, they must both be isomorphic to $\Bbb Z_4$, and hence isomorphic to eachother. $\endgroup$ – JMoravitz Jun 13 '16 at 2:15
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Hint: For the second group, use the order $1,i,-1,-i$.

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They are both isomorphic to they cyclic group of order 4, with $f$ as a generator of the first group, and $i$ a generator of the group of complex numbers.

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