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.
[Math] Methods to prove that two groups are isomorphic
abstract-algebragroup-theorysoft-question
Related Solutions
You can use the fundamental theorem of finitely generated abelian groups, which states that each of your groups can be expressed uniquely (up to isomorphism) as the product cycles of order the power of primes and compare factors and the prime powers of the cycles of any two such decompositions to determine whether they are isomorphic, up to the order of their factors.
In doing so, recall that $\;\mathbb Z_{m} \times \mathbb Z_n\;$ is cyclic and $$\;\mathbb Z_{mn} \cong \mathbb Z_m\times \mathbb Z_n \;\;\text{if and only if}\;\; \gcd(m, n) = 1$$
For example, let's take $\mathbb Z_8 \times \mathbb Z_{10} \times \mathbb Z_{24}$, and decomposing this gives us:
$$\mathbb Z_8 \times \color{blue}{\bf \mathbb Z_{10}} \times \color{red}{\bf \mathbb Z_{24}} \quad \cong \quad \mathbb Z_8 \times \color{blue}{\bf (\mathbb Z_{2} \times \mathbb Z_5)} \times \color{red}{\bf (\mathbb Z_{3}\times \mathbb Z_{8})} \cong {\bf \mathbb Z_{2} \times \mathbb Z_{3}\times \mathbb Z_5 \times \mathbb Z_{8}^2}\tag{1}$$
Now compare this to the decomposition you get for the expression you are to compare with $(1)$: $$\Bbb Z_4 \times \Bbb Z_{12} \times \Bbb Z_{40} \quad \cong \quad \mathbb Z_4 \times (\mathbb Z_3 \times \mathbb Z_4)\times (\mathbb Z_8\times \mathbb Z_5) \cong {\bf \mathbb Z_3 \times \mathbb Z_4^2 \times \mathbb Z_5 \times \mathbb Z_8}\tag{2}$$
Note that there the cyclic factors of $(1), (2)$ are not equivalent, up to the order of their appearance, hence not isomorphic.
Mathematicians typically don't use the word "homomorphic" that way, for the reason (as pointed out above) that for any two groups, $G,H$, the $0$-map (or trivial map):
$0:G \to H$ given by $0(g) = e_H$ for all $g \in G$ (map everything in $G$ to the group identity of $H$), always exists.
In any case, the existence, or non-existence, of a (non-trivial) homomorphism between two groups is a poor yardstick of similarity; for distinct primes $p, q$ there is no non-trivial group homomorphism $f: \Bbb Z/p\Bbb Z \to \Bbb Z/q\Bbb Z$, but very few mathmeticans would claim this groups are "dissimilar" (they are both simple cyclic groups).
It is (perhaps) more fruitful to think of a homomorphism $f: G \to f(G) \subseteq H$ as "preserving" partial information about $G$ "inside" of $H$. Some information about $G$ can be lost, as homomorphisms are not always one-to-one (when they are one-to-one, this is a highly desirable situation, as we have a "copy" of $G$ inside $H$, and $H$ may be "easier" to understand than $G$).
Groups can be very different from one another (as opposed to, say, finite-dimensional vector spaces over the reals, which are all rather alike). This is mostly because groups have so few rules, that some wildly disparate sets and operations can still qualify to be groups. Mathematicians have developed classifications, such as order, simplicity, etc. to help distinguish groups by type, and it turns out this is....complicated.
Best Answer
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.