3

How many non isomorphic groups are there

1) that have 2 elements

2) that have 3 elements

My solution:

1) There must be a identity element in a group and for each element $x$ there also has to be $x^{-1}$. If we look at 2 element groups, one of the elements is identity element and the other one has to have its inverse. Therefore there are no 2 element groups.

2) As a group doesn't have to be commutative, there's quite a lot of non isomorphic groups. I would appreciate some explanation on how to evaluate this problem without drawing graphs and simply counting non isomorphisms.

Rasmus Erlemann
  • 2,983

3 Answers3

3

(1) There is no reason $x$ can't be it's own inverse, such elements are called involutions.

Hint:

Consider $\{0,1\}$ with addition modulo $2$.

(2) You can characterize groups by their 'multiplication table'; how many ways are there to fill out a $3\times3$ table, provided you respect rules like $e* x = x$ for all $x$, etc.?

BaronVT
  • 13,613
2

1) One group $G=\{e,a\}$ with $a^2=a.a=e$ ($a^{-1}=a$)..

2) One group $G=\{e,a,a^2\}$ wiht $a^3=e$..

Hamou
  • 6,745
2

The identity $1$ is always the inverse of itself. So in a group with two elements there must be another element, say $a$. Since this must have an inverse, we must have $aa=1$. Thus the multiplication table is $$ 11=1,\quad 1a=a,\quad a1=a,\quad aa=1 $$ Does this table define a group?

More generally, let $p$ be a prime number and suppose $G$ is a group with $p$ elements. Since $p>1$ we can choose $g\in G$, $g\ne 1$. Then $$ \langle g\rangle=\{g^n:n\in\mathbb{Z}\} $$ is a subgroup of $G$. What does Lagrange's theorem imply at this point?

egreg
  • 238,574