Listing methods to prove that two groups are not isomorphic Topic: listing methods to prove that two groups are not isomorphic.
I understand that by nature, my question won't have one answer but more a list of answers. However I think that it is interesting to list properties that can be used to prove that two groups are not isomorphic.
Starting with some that come on top of my head:


*

*Using cardinality: the two groups have different cardinals.

*Using order of elements: one group has an element of a given order and not the second one.

*More generally using order of subgroups

*Using universal properties like commutativity

 A: 

*The automorphism group $Aut(G)$

*The cohomology groups $H^n(G,M)$ for $G$-modules $M$

*The homology groups $H_n(G,M)$ for $G$-modules $M$

*Solvability, Supersolvability, Nilpotency
A: Here are two more:


*

*Number of generators. To be more precise: you can prove that two groups are not isomorphic by proving that one of them is spanned by a set with a certain cardinal, whereas no set with that cardinal spans the other one.

*Quotient groups: if $G$ and $H$ are isomorphic and $N$ is a normal subgroup of $G$, then there has to be a normal subgroup $M$ of $H$ such that $G/N\simeq H/M$.

A: Here's another one : if for groups $G,H$, $G\simeq H$ if and only if $G-Set \simeq H-Set$ (equivalence of categories -though of course only the easy implication is needed here). This suggests that we may want to look at how $G$ actions behave comparatively to those of $H$. 
So a method would be : find a $G$-action that can't possibly be an $H$-action. I don't know if that often comes up though... 
Other methods include looking at the abelianization for nonabelian groups, or similarly at the derived subgroup. One may also want to look at the sizes of conjugacy classes. 
One may also look at first order notions, such as divisibility.
Let me now give not other methods but rather examples where those methods may be used, where I don't know of any easier method: 
-Looking at the center: this yields that $GL_n(\mathbb{R})$ and $GL_n(\mathbb{C})$ are never isomorphic for $n\geq 1$ (assuming you already know the $n=1$ case)
-Looking at orders of subgroups: this yields that for a field of characteristic $\neq 2$, $GL_n(K)\simeq GL_m(K)$ only if $n=m$ (look at the subgroups of exponent $2$: what are their sizes ?)
-Looking at the sizes of conjugacy classes: if I recall correctly, this implies that $\mathfrak{S}X$ and $\mathfrak{S}Y$ can only be isomorphic if $X$ and $Y$ have the same cardinal (or one of them is empty and the other one is a singleton) (this is of course trivial for finite $X,Y$, by looking at the cardinality, but not so much for infinite $X,Y$). Indeed, if one looks at the smallest size of conjugacy class of any element $\neq 1$, one finds $|X|$, for infinite $X$ (see here)
-Number of generators: $(\mathbb{Z}/p\mathbb{Z})^2$ is not isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$
A: Basically any property that is preserved by group isomorphisms will do. This includes:


*

*order of elements

*commutativity

*amount of subgroups of a certain (finite) order

*amount of Sylow p-subgroups

*cardinality of the group

*being cyclic (2 cyclic groups of same cardinality are isomorphic)

*order of subgroups



Another method (for finite groups) is to look at character tables. If two groups have different character tables, they can't be isomorphic. Notice that the converse is false: two groups can have the same character table without being isomorphic (If I remember correctly $D_8$ and $Q_8$ are a counterexample when considering the character table for the complex irreducibel representations, but correct me if I'm wrong!)

In practice, one can look at special subgroups like the center and normalisator of a group, as these are often easier to understand than the entire group. 
