Question

The group ${{Z}_{4}}$ under addition modulo 4 has
(a) Exactly one proper subgroup
(b) Only two proper subgroups
(c) No subgroups
(d) Infinitely many proper subgroups

Answer 1

Hint: Here, we will use the concept of addition modulo on the elements of ${{Z}_{4}}$ and check whether they form a subgroup or not. Any group is called a cyclic group if all its elements can be generated using only one of its elements. A group is called a subgroup of a cyclic group ${{Z}_{n}}$ if the HCF of a and n is 1, here denotes a cyclic group whose generator is a. The HCF of two numbers ‘a’ and ‘b’ is the greatest number which divides both of them.

Complete step-by-step answer:
By addition or multiplication modulo we mean that we take and fix any positive integer ‘n’ greater than 1. Then to add or multiply any two integers modulo n just add or multiply them as usual and divide the result by n and take the remainder ( integer division where you get a integer quotient and an integer remainder lying in the set {0, 1, 2,…..n-1} ). This remainder is termed as the sum or product, as the case may be, of the two integers you started with, modulo n.
Now, since we know that ${{Z}_{4}}=\left\{ \left[ 0 \right],\left[ 1 \right],\left[ 2 \right],\left[ 3 \right] \right\}$, where $\left[ 0 \right],\left[ 1 \right],\left[ 2 \right]\text{ and }\left[ 3 \right]$ are residue classes.
Now,
$\begin{align}
  & \left[ 1 \right]+\left[ 1 \right]=2\left( \bmod 4 \right)=\left[ 2 \right] \\
 & \left[ 1 \right]+\left[ 1 \right]+\left[ 1 \right]=3\left( \bmod 4 \right)=\left[ 3 \right] \\
 & \left[ 1 \right]+\left[ 1 \right]+\left[ 1 \right]+\left[ 1 \right]=4\left( \bmod 4 \right)=\left[ 0 \right] \\
\end{align}$
Thus, we see that all the elements of ${{Z}_{4}}$ can be generated from $\left[ 1 \right]$. So, ${{Z}_{4}}$ is a cyclic group.
Since, a group is said to be a subgroup of ${{Z}_{n}}$ only if HCF of a and n = 1.
The elements belonging to ${{Z}_{4}}$ are $\left[ 0 \right],\left[ 1 \right],\left[ 2 \right]\text{ and }\left[ 3 \right]$.
Among these 4 elements, the numbers whose HCF with 4 is 1 will be called a subgroup of ${{Z}_{4}}$.
Since, HCF of 1 and 4 is 1, so $\left[ 1 \right]$ is a subgroup of ${{Z}_{4}}$.
HCF of 2 and 4 is 2, which is again not equal to 1. So, $\left[ 2 \right]$ is not a subgroup of ${{Z}_{4}}$.
HCF of 3 and 4 is 1, so $\left[ 3 \right]$ is a subgroup of ${{Z}_{4}}$.
Hence, option (b) is the correct answer.

Note: A possibility of mistake here is that while checking whether ${{Z}_{4}}$ is a cyclic group, we add $\left[ 1 \right]$ four times to get $\left[ 4 \right]$. Now, we have to always take the remainder obtained after addition and that is why
$\left[ 4 \right]=\left[ 0 \right]$ because on dividing 4 by 4, a zero remainder is obtained. It should be noted that HCF is always the greatest number that divides both the numbers.