So my textbook says that there are two classifications of groups of order four. Those two are:
$\mathbb{Z}_4\cong\{0,1,2,3\}$ under $+_4$, and
$\mathbb{K}_4\cong$ symmetry group of a (non-square) rectangle.
It also says if $P\cong Q$, and $P$ has $k$ elements of order $n$, then $Q$ has $k$ elements of order $n$. And that groups of order 8 or less can be classified entirely by the orders of their elements.
So, let $(O,N)$ describe a group, $A$, such that $O$ represents an order (where $O$ divides $|A|$) and $N$ represents the number of elements (in the underlying set of $A$) with that respective order $O$.
Let $G$ be a group such that $|G|=4$.
$G$ can only be broken down in one of the following ways the following ways:
$G_1=(1,1),(2,3)$
$G_2=(1,1),(2,2),(4,1)$
$G_3=(1,1),(2,1),(4,2)$
$G_4=(1,1),(4,3)$
Since one and only one element can have an order of one, and the other three elements can either have an order of two or four (ignoring order).
Given this, $\mathbb{Z}_4\cong G_3$, and $\mathbb{K}_4\cong G_1$, leaving both $G_2$ and $G_4$ without a group to isomorphise with.
I've also seen online that a cyclic group has exactly one generating element, whereas my textbook says that a cyclic group has at least one generating element. I feel like a clarification there might clear this up. Or is it the case that $G_2$ and $G_4$ aren't possible? If so, how would I prove that (for larger orders)?
Best Answer
$G_2$ is impossible. If $|G|=4$ and $x$ has order $4$, then $x^{-1}$ has order $4$ and is distinct from $x$. So we cannot have $(4,1)$.
$G_4$ is impossible. If $|G|=4$ and $x$ has order $4$, then $x^2$ has order $2$. And if there is no element of order $4$ then every non-identity element in $G$ has order $2$ by Lagrange's Theorem. So we cannot have $(2,0)$.
Finally, it is certainly not the case that a cyclic group has a unique generating element. Indeed, in a cyclic group of prime order, every non-identity element is a generator. (Perhaps you can provide a source where you saw this?)