Methods to prove that two groups are isomorphic

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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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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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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 $H$.