A notational question about Cyclic Groups

Question

A well-known result about Cyclic Groups is:

Let $G = \langle a \rangle$ be a cyclic group of order $n$. Then, for every $k \in \mathbb{Z}$: \begin{align} G = \langle a^k \rangle \Leftrightarrow \gcd(n,k) = 1. \end{align}

In order to illustrate this result my textbook considers the cyclic group $\mathbb{Z}/12\mathbb{Z} = \{\overline{0}, \overline{1}, \ldots, \overline{11}\}$ and states that the generators of this group are $\overline{1}, \overline{5}, \overline{7}, \overline{11}$.

My question now is: How do I have to understand the expression $a^k$?

Usually, I would read it as $a^k = \underbrace{a \cdot a \cdot \ldots \cdot a}_{k-\text{times}}$, but in the context of the above mentioned example it would rather be $a^k = \underbrace{a + a + \ldots + a}_{k-\text{times}}$. (Using "$\cdot$" would not make much sense since $\overline{1}$ for example is no generator.)

Since I confused myself a little bit about this, there is a more general question: If we are talking about Cyclic Groups do I always have to think about "$+$" as the operation instead of "$\cdot$"? Just to make sure: When I am talking about $(\mathbb{Z}/n\mathbb{Z})^{\times}$ (the units of $(\mathbb{Z}/n\mathbb{Z})$) this is always a group with "$\cdot$", right?

Answer 1

If $(G, *)$ is a group and $a \in G$, then $a^k$ simply means $\underbrace{a*\cdots * a}_{k \text{ times}}$ for $k > 0$.

If you are denoting the operation as $+$, then you have $+$ instead of $*$ above.

So it simply is a matter of what symbol you use for the operation.

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Now, given an abelian group, it is a common convention to use $+$ for the group operation since it's commutative. In this case, one may also sometimes write $ka$ instead of $a^k$.
So, if your group is cyclic, then it indeed abelian and you may identify it with $\Bbb Z/n\Bbb Z$ and use $+$.

However, one also has another very natural cyclic group which is written multiplicatively, namely, the group of $n$-th complex roots of unity.
This is the group $\{1, \zeta, \ldots, \zeta^{n - 1}\}$, where $\zeta = e^{2\pi\iota/n}.$

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To summarise: Depending on your symbol for the group operation, you have the appropriate definition of $a^k$.
So you are correct that you would think of "$\cdot$" when thinking about $(\Bbb Z/n\Bbb Z)^\times$.

Answer 2

It is very easy to get confused in group theory by according more "significance" to the notation used than it deserves.

All that is relevant about an operator is how it operates on the group elements.

There is a danger in using conventional notations, e.g. $\cdot$ or $+$, because they can lead the student down the road of thinking: "This means multiplication", or "This means addition", whereas in fact this may not be the case.

It is a common convention to use $+$ as a general operator in an abelian group, that is, one in which the operator is commutative. And it is often the case that $+$ does actually mean "addition" of some kind or other, for example, modulo addition or matrix addition.

Similarly for symbols that make you think of "multiplication", that is, $\times$ or $\cdot$.

Hence the confusion between the notation $a^k$. It means:

$$a^k = \begin{cases} e & : k = 0 \\ a \circ a^{k - 1} & : k > 0 \\ \left({a^{-k} }\right)^{-1} & : k < 0 \end{cases}$$

where $\circ$ denotes the operator, $e$ denotes the identity and $^{-1}$ denotes the inverse.

This is the case whatever the operator is.

That means, when the symbol you've picked for your operator happens to be $+$, you have the bizarrely anti-intuitive notation: $a^k = a + a + \cdots + a$ (where there are $k$ instances of $a$). And you say to yourself: that should really be $k \cdot a$ or $k \times a$ or $k a$ or whatever.

Okay, yes, maybe that "makes more sense" to you. In that case, you can amend your notation again, to present your definition as:

$$k \cdot a = \begin{cases} 0 & : k = 0 \\ a + ((k - 1) \cdot a) & : k > 0 \\ -(-k \cdot a) & : k < 0 \end{cases}$$

where $+$ denotes the operator, $0$ denotes the identity and $-$ denotes the inverse.

But now you have to remember that $k$ is an integer, while $a$ is not necessarily an integer because it is an element of whatever your group is. And so now you have the same notation for your operations on your group as you have on your operations on your integers.

And this can make it a lot more confusing.