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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)?

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

1 Answer 1

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$G_2$ 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?)

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  • $\begingroup$ Thanks. I think I may have read "can be generated my a single element" as "only a single element can generate it". As for the number of generators of a cylic group, I understand now that there must be an even number of them (for groups of orders greater than 2). But doesn't that mean that for groups with an order n (which is larger than 4) there can be several cyclic groups that aren't isomorphic with each other. E.g. one has (n,2) (two generators) and the another has (n,4) (four generators). Since my book seems to imply that there can only be one cyclic group of each order (up to 8 at least) $\endgroup$
    – Chris Whitehead
    Jul 28, 2020 at 16:10
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    $\begingroup$ 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. $\endgroup$
    – halrankard
    Jul 28, 2020 at 16:16
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    $\begingroup$ See Theorem 9.8 of these notes: math.purdue.edu/~arapura/preprints/algebra9.pdf $\endgroup$
    – halrankard
    Jul 28, 2020 at 16:19

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