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What are some of the more effective methods to prove that 2 groups are isomorphic. The methods that I am currently using now is to always find a function from A to B that is bijective and homomorphic which is rather tedious. However, I believe that there should be more effective ways other than constructing a function all the time. Can someone share with me some of the more effective methods? In one of my assignment question that asks me to show that $Q_8/Z(Q)$ is isomorphic to klein four group. The solution given is that other than the identity, all the elements of $Q_8/Z(Q)$ and klein four groups are of order two. So they are isomorphic. I am wondering why such a conclusion can be made. Thanks.

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  • $\begingroup$ How big is $Q_8 / Z(Q)$? How many groups are there of that order? What do their elements look like? $\endgroup$ – user171177 Nov 4 '14 at 14:07
  • $\begingroup$ Have you gone over the classification if groups of small order? Order 4 only had the cyclic group of order 4, which doesn't work, and the Klein 4 group. $\endgroup$ – Matt Samuel Nov 4 '14 at 14:07
  • $\begingroup$ nope this is my very first course on abstract algebra $\endgroup$ – user10024395 Nov 5 '14 at 9:28
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That conclusion that $Q/Z(Q) \cong V$ is founded on that $|Q/Z(Q)| = 4$. Thus $Q/Z(Q)$ is isomorphic to either $\mathbb Z_4$ or $V$. (Why? These group operations are the only valid ones on a set of four elements. To see this, try writing out a group table for a group of four elements just from the axioms. You will end up with exactly two possibilities up to isomorphism)

So our task is to figure out which group $Q/Z(Q)$ is isomorphic to. We know that order is preserved through isomorphism, and $\mathbb Z_4$ has two elements of order 4 while $V$ has none. Thus observing that $Q/Z(Q)$ has all elements of order $\le 2$ shows that it cannot be isomorphic to $\mathbb Z_4$ and we are done.

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  • $\begingroup$ Here's a handy list list of small groups. $\endgroup$ – Nikolaj-K Nov 4 '14 at 14:32
  • $\begingroup$ why "these group operations are the only valid ones on a set of four elements"? $\endgroup$ – user10024395 Nov 5 '14 at 9:33
  • $\begingroup$ what happens if there are more than 2 possibilities, do I have to eliminate them one by one? Isn't that very inefficient? $\endgroup$ – user10024395 Nov 5 '14 at 9:35
  • $\begingroup$ For an example of why these are the only groups on four elements, see, for instance, Fraleigh's "A First Course in Abstract Algebra", Chapter 4, Exercise 20. And if you were to use this method, then yes, one would have to eliminate them one by one. This method only works well for small-order groups. $\endgroup$ – Joshua Mundinger Nov 5 '14 at 13:35
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I will answer the general question first. Often groups have a known presentation, i.e. a description by generators and relations between them (such that every other relation between the generators can be deduced from them). From an abstract point of view, a presentation is really a universal property: For example, $D_n = \langle a,b : a^n=b^2=(ab)^2=1\rangle$ means that if $G$ is any group and $x,y \in G$ are elements with $x^n=y^2=(xy)^2=1$ in $G$, then there is a unique homomorphism $D_n \to G$ such that $a \mapsto x$ and $b \mapsto y$. Universal properties allow us to construct homomorphisms, and also inverse homomorphisms if the target has a presentation. This way you can test if two groups, with given presentations, are isomorphic.

Now about the example: The quaternion group is $Q = \langle i,j : i^4=1, i^2=j^2, j^{-1} i j = i^{-1} \rangle$. Its center is $\langle i^2 \rangle$. Hence, $Q/Z(Q) = \langle i,j : i^2=j^2=1, j^{-1} i j = i^{-1} \rangle = \langle i,j : i^2=j^2=1, ij=ji \rangle$. This is exactly $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, since $\mathbb{Z}/2\mathbb{Z} = \langle a : a^2=1 \rangle$ and in general a presentation of $G \times H$ can be produced by "joining" presentations of $G$ and $H$ and adding the relation that the generators of $G$ commute with the generators of $H$.

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