# Showing that $\{0, 1\}$ is a group under addition modulo $2$

I'm considering a set $G = \{0,1\}$ under addition modulo 2. I.e. $$a*b = a + b\bmod 2, \quad \quad \forall \ a,b \in G.$$ I am able to show that there exists an identity element, $0$.

Showing that $*$ is assosiative and that there exists an inverse, is more difficult to me.

Can someone help me to show this?

• Concerning inverse, compute $a\ast a$. What is the inverse of $a$ ? – Dietrich Burde Jan 7 '15 at 21:41

Showing that the identity $e \in G$ exists means showing $a= a\star e = e \star a$ for all $a \in G$, and as you correctly pointed out $e = 0$. Next I would prove every element has an identity, since you only have one non-identity element left in $G$ to consider. By definition, $$1 \star 1 = 1+1\bmod 2 \\ = 2\bmod 2 \\ = 0$$ So $1$ is its own identity (making it an idempotent element, if that matters much to you). Now for associativity, let $a,b,c \in G$ and consider $a \star b \star c$. Since you have three elements $(a,b,c)$ that can be valued as either $1$ or $0$, you could run through by hand all $2^3=8$ possible operations of the form $a\star b \star c$ and show $(a \star b) \star c = a \star (b \star c)$ in all eight cases. Otherwise you can take the more general approach and try: $$(a \star b) \star c = (a+b\bmod 2) \star c \\ = (a+b\bmod 2)+c\bmod 2$$ compared to $$a \star (b \star c) = a\star (b+c\bmod 2) \\ = a+ (b+c\bmod 2)\bmod 2$$ Now can you show that $$(a+b\bmod 2)+c\bmod 2 = a+ (b+c\bmod 2)\bmod 2$$ ?

Since $G=\{0,1\}$, you have just eight identities to check for associativity:

$$0*(0*0)=(0*0)*0\\ 0*(0*1)=(0*0)*1\\ 0*(1*0)=(0*1)*0\\ 0*(1*1)=(0*1)*1\\ 1*(0*0)=(1*0)*0\\ 1*(0*1)=(1*0)*1\\ 1*(1*0)=(1*1)*0\\ 1*(1*1)=(1*1)*1$$

The inverse of $0$ is of course $0$. What can the inverse of $1$ be? Is it?

• Also, the operation is commutative, so you need to check only half of the cases. – sas Jan 7 '15 at 22:37
• @sas Not really. – egreg Jan 7 '15 at 22:39
• In your list you need check cases number $2, 4, 5, 7$. – sas Jan 7 '15 at 22:41
• @sas Choosing the ones to check takes longer than the complete check. But in more complex cases it can help. – egreg Jan 7 '15 at 22:42

For all $a,b \in G$
$$(a*0)*b=a*b,\quad a*(0*b)=a*b$$
and
$$(a*1)*b=(1-a)*b=1-a+b \bmod 2 = \begin{cases} 1, \quad\text{if}\quad a=b \\ 0,\quad\text{if}\quad a\neq b \end{cases}$$
$$a*(1*b)=a*(1-b)=a+1-b \bmod 2 = \begin{cases} 1, \quad\text{if}\quad a=b \\ 0,\quad\text{if}\quad a\neq b \end{cases}$$

So, for all $a,b,c\in G$ it is true that $$(a*c)*b=a*(c*b).$$