Show that an abelian group $G$ of order 55 must be cyclic.

Question

I know that in order to be cyclic: A group G is called cyclic if there exists an element g in G such that G = ⟨g⟩ = { $g^n$ | n is an integer } by wikipedia. But I just get lost in how simple it looks in terms of if we let n=1, then generate everything right?

Answer 1

You can't let $n$ be something, that $n$ is part of the description of the group: $\{g^n | n \text{ an integer}\}$ is notation for the set

$$ \{ ..., g^{-3}, g^{-2}, g^{-1}, e, g, g^2, g^3, ... \}.$$

That means the set of all elements of the form $g^n$, not just one particular $n$. It's generated by $g$, not just any $g^n$.

You have to show that if a group is abelian and has $55$ elements, it has to be of that form, i.e. there has to be an element of the group such that every other element is a power of it.

One way to do it is to show that since $55=11\cdot 5$, there has to be an element $g$ of order $11$, and another element $h$ of order 5. Then show that $k=gh$ has to have order $55$, and so $\{k^n | n\text{ is an integer}\}$ is the entire group.

It's not hard to show $k^{55}$ has to be the identity element, so one has to show $k^r$ is the identity only if $r$ is a multiple of $55$. Suppose $k^r=e$, then we have $e= k^r=(gh)^r = g^r h^r$ since the group is abelian, which means $g^r = h^{-r}$. Now $g^r$ is in the cyclic group generated by $g$, and $h^{-r}$ is in the group generated by $h$. The element $g^r=h^{-r}$ must then belong to both $\langle g \rangle$ and $\langle h\rangle$. By considering the orders of these groups, you can show this element must be the identity itself. This would then imply that $r$ is divisible both by $5$ and by $11$.

Answer 2

Your idea doesn't make sense. A group is cyclic if there exists an element $g\in G$ for which every element $h\in G$ has $h= g^n$ for some number $n$. You don't get to pick $n$ and you don't get to just decide that it is 1; it will be different depending on which element $h$ you are trying to make from $g^n$.

Consider for example the group consisting of $\{1, 2, 4, 8\}$ with multiplication taken modulo 15. The group is generated by the element $2$, because: $$\begin{align} 1 & = 2^4\pmod{15}\\ 2 & = 2^1\pmod{15}\\ 4 & = 2^2\pmod{15}\\ 8 & = 2^3\pmod{15}\\ \end{align}$$

with $n$ here being $4, 1, 2, $ or $3$ repsectively. (The group is also generated by $8$, but not by $4$.)

If you wanted to use this, the proof would look like this:

1. You are going to show that there is some $g$, so that for every $h\in G$, there is $n$ for which $g^n = h$.
2. You figure out one specific element $g$ which you claim generates $G$.
3. You suppose that someone given you an element $h\in G$.
4. You show how to find $n$ for which $g^n = h$, you can choose a different $n$ depending on what $h$ is, but you don't get to change $g$.

Picking $n=1$ won't work, because in step 3 the other person might give you some $h$ that is not equal to the $g$ you picked in step 2. The $n$ you pick in step 4 will depend on the $h$ you are given in step 3.

Do you know the classification theorem for abelian groups, the one that says that any abelian group must factor into a direct product $\Bbb Z_{p^i}\times \Bbb Z_{q^j}\times\cdots$ where $p, q\ldots $ are prime? That is what I would use to solve this exercise.

Answer 3

The order of any element of the group divides $55$. So the order of an element $\ne e$ is either $5$, $11$, or $55$. We'll show that there are elements of order $55$. Any such element will generate the group and hence the group is cyclic.

First, a basic fact. Let $p$ a prime number and $G$ a finite abelian group such that $g^p=e$ for all $g \in G$. Then the order of $G$ is a power of $p$. Indeed, $G$ can be considered as a vector spaces over the field $\mathbb{Z}/p$. Let $d = \dim_{\mathbb{Z}/p} G$. Then $|G|= p^d$.

Using the above result and the previous observation it follows that there exists an element of order $55$ ( we are done then) or there exist two elements $g$, $h$ of orders $5$ and $11$ respectively. We claim that the order of $gh$ is $55$. Indeed $(gh)^{55} = g^{55} \cdot h^{55} = e\cdot e = e$ so its order divides $55$. It cannot be $1$ since $g \ne h^{-1}$, being of different orders, and it's neither $5$ nor $11$. For instance $$(gh)^{11}= g^{11} \cdot h^{11} = g \ne e$$

To show that the hypothesis "abelian" is necessary we present an example of a non-abelian group of order $55$ (the other group of order $55$) :

$G$ is a group of the affine transformations of $\mathbb{Z}/11$,

$$T_{a,b} \colon x \mapsto a\cdot x + b$$

where $b \in \mathbb{Z}/11$ and $a \in (\mathbb{Z}/11^{\times})^2= \{1,3,4,5,9\} \ \text{mod} 11$

For these kind of examples we used $5 | (11-1)$. It turns out that $\it{any}$ group of order $p\cdot q$ where $p<q$ primes and $p \not | (q-1)$ is cyclic.

Answer 4

The abelian groups of size 55 up to isomorphism are just the single direct product Z5 X Z11 since that is the only way that 55 can be factored into primes. Since 5 and 11 are relatively prime Z5 X Z11 is cyclic and hence all abelian groups of size 55 are cyclic.