# Only Two Isomorphism Classes of Groups of Order Four

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)?

• Title: There is only one classification of groups of order $4$. It says that either $G\cong C_4$ or $G\cong C_2\times C_2$. Jul 28, 2020 at 15:36
• Which textbook are you referring to? Jul 28, 2020 at 15:36
• If $g$ generates a cyclic group $G$, then its inverse $g^{-1}$ generates $G$. If the order of $G$ is greater than $2$, then $g\ne g^{-1}$.. Jul 28, 2020 at 15:37
• I've edited the title, but now I see Dietrich Burde's suggestion so feel free to change it if you like that better. Jul 28, 2020 at 15:39
• @Shaun it's the Edexcel A-Levels Further Pure 2, chapter 2 Jul 28, 2020 at 15:52

$$G_2$$ 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 $$(2,0)$$.
• A priori what you say makes sense. But it is a theorem that for any finite $n$ there is a unique cyclic group of order $n$. One will always have $\varphi(n)$ generators in a/the cyclic group of order $n$ (i.e., it is always $(n,\varphi(n))$ in your schematic). Here $\varphi$ is Euler's totient function. Jul 28, 2020 at 16:16