In my recently appeared test I got a question mentioned below:
1) In an additive group $G$ of integers, the order of inverse element $a^{-1}$, for all $a \in G$ is
(a) $0$
(b) $1$
(c) $\infty$
(d) none of these.
I got it right by choosing option (c). But I was in confusion as I thought element of order zero and infinity are the same. does element of order zero and element of order infinity in a group mean the same? Is option (a) also correct?
No, the order is infinity, not zero. I have a few more details that might help:
Note: I will use the notation $na$ to mean $\underbrace{a+a+\cdots+a}_{n\ \mathrm{times}}$.
Let $G$ be an additive group, $a \in G$. The order of $a$, $|a|$, is defined to be the smallest $n \in \mathbb{N}$ such that $na = 0$. If such an $n$ does not exist, $|a|$ is infinite.
However, if we have a ring $R$, the characteristic of $R$ is the smallest $n\in\mathbb{N}$ such that for all $r\in R$, $nr = 0$. It's essentially the least common multiple of the orders of each of the elements in $R$. However, if such an $n$ doesn't exist, we don't say the characteristic is infinite. Instead, we say $Char(R) = 0$.
This is probably where your zero/infinity confusion comes from.