Explanation: Calculation Rules in a Group with Additive Notation

(related to Motivation: Calculations in a Group)

For someone who is new to group theory but is familiar with arithmetics it might be surprising, that in a group $(G,\ast)$, the same calculation rules apply if we exchange the operation $\ast$ by an addition sign $(G, +)$ or by a multiplication sign $(G, \cdot).$

For instance, in the additive notation, the above calculation rules in a group can be formulated as follows:

  1. $-e=e.$
  2. $-(-x)=x$ for all \(x\in G\).
  3. For all \(a,b\in G\), if \(a + x=b\) then \(x=-a+b\).
  4. For all \(a,b\in G\), if \(x + a=b\) then \(x=b-a\).
  5. If $a+x=a+y$ then $x=y$ and if $x+a=y+a$ then $x=y.$
  6. $-(x+y)=-y+(-x)$ for all \(x,y\in G\).
  7. \(x_1 + x_2 + x_3 + \ldots + x_n:=(\ldots((x_1 + x_2) + x_3) + \ldots + x_n\) holds for all \(x_i\in G\).
  8. \(x_{k_1} + x_{k_2} + \ldots + x_{k_n}=x_{1} + x_{2} + \ldots + x_{n}\)

The corresponding proof of these (seemingly different) rules is exactly the same except for the notation. Please try to write down the proof in the additive notation it as an exercise!

Similarly, while the exponentiation in a group is a new operation, which can be defined, if you use the multiplicative notation $(G,\ast)$, the same can be done in the additive case $(G, + ).$ Here, instead of an "exponentiation", we can define a "multiplication" as a new operation and explain it in terms of addition. To be more concrete, for an integer $n\in \mathbb Z$, we can define:

\[n\cdot x := \begin{cases} e & \text{ if } n=0 \\ x + (n-1)\cdot x & \text{ if } n > 0. \end{cases}\]

And for negative integers $-n$:

$$-n\cdot x:=n\cdot (-x).$$

But you have to be very cautious!

Note that the dot sign "$\cdot$" in $n\cdot x$ does not (!) mean that we multiply the integer $n$ by a group element $x$. It only means that we introduce a shorter notation for adding $n$ times the element $x\in G$ to each other. That is also the reason why it makes sense to set the result to the neutral element $e\in G$ in the case $n=0,$ since we add "none" to each other.

For the sake of completeness, we want to reformulate the exponentiation rules and translate them into the additive notation for groups:

\[\begin{array}{cl} (1)&n\cdot x+ m\cdot x=(n+m)\cdot x,\\ \end{array} \]

If the group is Abelian, then \[\begin{array}{cl} (2)&m\cdot (n\cdot x)=(mn)\cdot x.\\ (3)&n\cdot x+n\cdot y=n\cdot (x + y). \end{array} \]

Again, the proof would be exactly the same (try it as an exercise!).

Please note that the exponentiation rules - written down in the additive notation - are strongly reminiscent of a kind of a "distributivity rule" for groups.

But again, be cautious!

There is no distributivity rule for groups since in a group only one operation is defined - but for distributivity, we need two: addition and multiplication. Remember, and this is only a repetition of what has been said above: $n\cdot x$ does not mean that we multiply an integer $n$ by the group element $x.$ It only means that we add $x\in G$ $n$ times to each other.

The reason, why we haven't chosen the additive notation for introducing these calculation rules is that if you are a beginning student of group theory, there is a danger of confusing the operation sign "$\cdot$" in $n\cdot x$ with the group operation, which is only the addition $(G, +).$

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