Exercise 1

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Groups

Ex1: Show that a group has precisely one identity element

To be an identity element for a group we have to fulfil the following two rules:

$$a*e = e * a = a$$ $$a*a^{-1} = e$$

If we assume we have to identity elements $e$ and $e’$, then to fulfil above rule it follows that

$$e*a = e'*a = a$$ $$e*a*a^{-1} = e'*a*a^{-1}$$ $$e*e = e'*e$$ $$e = e'$$ $$\text{Contradiction, if there are two identity elements, they have to be the same element}$$

Ex2: Show taht for any element $s$ in a group, there is precisely one inverse to $s$.

Ex3: Create the addition and multiplication tables modulo 4.

Ex4: Show that $G_4$ with addition modulo 4, $(G_4, +)$ is a group.