# Can we expand $a^b$ to where both are elements of a group? Deriving exponentiation from multiplication group

For some group $(G,\cdot)$ my professor defined $a^n$ where $a \in G$ and $n \in \mathbb{Z}$ to be $a^n=a \cdot a \cdot a ...a$ where there are $n$ copies of $a$ (if $n$ is a negative, it's $n$ copies of the inverse and if $n=0$ then the result is the identity element). I was wondering if there was some other definition out there of exponents where the exponent was also from the group.

So is there a definition for $a^b$ with $a, b \in G$? The result would hopefully be another element from the group, but how can you have $b$ repetitions if $b$ doesn't actually count anything?

## 1 Answer

You can define $a^b$ as whatever you want it to be, of course, but in order for it to be considered meaningful, it must be similar to "exponentiation" in some (precisely defined) way. There are different options for this.

You may wish this exponentiation to follow one or more of the usual "laws of exponents". For example, $(a^b)^c = a^{bc}$. Since there is no second operation $+$, we cannot meaningfully write $(a^b)(a^c) = a^{b+c}$ [unless we define some new operation].

One (very commonly used) definition that satisfies the first property is conjugation: For $a, b \in G$, define $a^b = b^{-1}ab$. Then $(a^b)^c = c^{-1}(a^b)c = c^{-1}b^{-1}abc = (bc)^{-1}a(bc) = a^{bc}$ [$\ne a^{cb}$, in general].

This can be generalized. For any fixed homomorphism $\varphi\colon G \to \operatorname{Aut}(G)$ from the group $G$ to its automorphism group $\operatorname{Aut}(G)$, we can define, for $a, b \in G$, $a^b = \varphi_b(a)$, where $\varphi_b$ denotes the automorphism $\varphi(b)$. But now, we will get $(a^b)^c = a^{cb}$. To avoid this, we could define $a^b = \varphi_b^{-1}(a)$ (though that is messy as a definition). The proof that these definitions work as claimed is left as an exercise to the reader! Note that all of this ties up very nicely with left and right actions.

If we wish to have both of the aforementioned properties of exponents, we need two operations, so consider a ring $(R, +, \times)$. Let $G$ be the additive group of the ring, written multiplicatively as $(G, \cdot)$, where $G = R$, and for $a, b \in G$, $ab = a +_R b$ (the sum of $a$ and $b$ in the ring $R$). We can then define $a^b$ as $a \times_r b$ (the product of $a$ and $b$ in the ring). Now, we have both properties! Note that all operations written in the exponent actually take place in the ring $R$. So, the product (in $G$), of $a^b$ and $a^c$ is $(a^b)(a^c) = (a \times_R b) +_R (a \times_R c) = a \times_R (b + c) = a^{b+c}$, where the $+$ in the exponent is addition in the ring. Also, $(a^b)^c = (a \times_R b) \times_R C = a \times_R (b \times_R c) = a^{b \times c}$, where the $\times$ in the exponent is multiplication in the ring. If $b$ is a unit of the ring, it has a multiplicative inverse $b^{-1}$, so we can even find the "$b$-th root of $a$"!

Needs some correction:

Now, maybe we wish for a different definition — one that "agrees" with the "integer exponent" $a^n$ already defined. But how can the two agree if the exponents are elements of two different sets (the integers vs. the group elements)? Let $M$ be a monoid that contains (a copy of) $\mathbb Z$ or $\mathbb Z_n$ as a subgroup (respectively as the exponent of $G$ is infinity or a finite number $n$). In other words, there is a monoid homomorphism from the monoid $(\mathbb Z, +)$ of (additive) integers into $M$, and the homomorphic image is the subgroup mentioned in the previous statement. Further, suppose this monoid $M$ acts on $G$ in such a way that for every (image of the) integer $m \in M$, the action of $m$ on $G$ is defined by $a \mapsto a^m$. Now, we can define a homomorphism $\varphi\colon G \to M$, from the monoid $G$ (every group is a monoid) into $M$, and as before, define $a^b = \varphi_b(a)$. If this homomorphism satisfies the property that $a^b = a^m$ whenever $\varphi(b) = m$ is an element of the submonoid $\mathbb Z$ or $\mathbb Z_n$ of $M$, we can say that our definition of exponentiation agrees with integer exponentiation.

• To fix your last comment, just define it as $bab^{-1}$ instead. – Tobias Kildetoft Apr 15 '16 at 7:24
• Indeed, if you use that definition, we get $(a^b)^c = a^{cb} \ne a^{bc}$ (intuitively less appealing). I've seen a "left exponentiation" notation used for that sometimes: ${}^b a = bab^{-1}$. – M. Vinay Apr 15 '16 at 7:26
• Woops, I actually read your comment wrong (too little coffee today yet). Indeed the definition you give here is the one with the nicest properties for this purpose. – Tobias Kildetoft Apr 15 '16 at 7:29
• Another way to formulate this is in terms of having another group $H$ act on the group $G$, for which this is a special case (and the usual exponentiation also being a special case). – Tobias Kildetoft Apr 15 '16 at 7:39
• What if we use a ring like $(R,+, \cdot )$? So that $(a^b)(a^c)=a^{b+c}$ can have some meaning? In this case, the exponential notation would still stem from the multiplication part of the ring – Cordello Apr 15 '16 at 7:42