Answer to Question #211951 in Abstract Algebra for jpm

Question

Cyclic Groups

1. Let Z denote the group of integers under addition. Is every subgroup of Z cyclic? Why? Describe all the subgroups of Z.
2. List the elements of the <i>, i.e. cyclic subgroup generated by i of the group C* of nonzero complex numbers under multiplication.
3. Let G=<a> and |a|=24. List all generators for the subgroup of order 8.
4. Draw the subgroup lattice diagram for Z₃₆ and U(12).

Answer 1

1.

Let $H$ be arbitrary subgroup of $\mathbb{Z}$ , $d\in H\setminus\{0\}$ an element of $H$ with the minimal absolute value. Then cyclic group is a subgroup and a subset of $H$ . In fact, $\langle d\rangle=H$ . Indeed, if $x\in H\setminus \langle d\rangle$, then one can divide $x$ by $d$ with remainder $r\ne 0$ , such that $|r|<|d|$ (applying Euclid's algorithm): $x=dq+r$ . Then $r=x-dq\in H$ and $|r|<|d|$ . This contradicts to the condition that $d$ is a non-zero element of $H$ with the minimal absolute value. Therefore, $H=\langle d\rangle$ , i.e. any subgroup of $\mathbb{Z}$ consists of all multiples of some integer $d$ .

2. List the elements of the $\langle i\rangle$ , i.e. cyclic subgroup generated by $i$ of the group C* of nonzero complex numbers under multiplication.

$i^0=1,\, i^1=i, \, i^2=-1,\, i^3=-i, i^4 =1$ .

Therefore, cyclic subgroup generated by $i$ of the group $C^*$ is $\{1, -1, i, -i\}$.

3. Let $G=\langle a\rangle$ and $|a|=24$ . List all generators for the subgroup of order $8$ . We may assume $G=Z_{24}$ without loss of generality.

Let $H\subset G$ be a subgroup of order 8. Since order of any element of $H$ divides order of $H$ , then $8x=0$ for all $x\in H$ . Solving the equation $8x=0\mod 24$ , we obtain $x=0 \mod 3$ . Therefore $H=\{0, 3, 6, 9, 12, 15, 18, 21\}$ and any odd element can be taken as a generator of $H$ . They are $\{3, 9, 15, 21\}$.

4. Draw the subgroup lattice diagram for Z₃₆ and U(12).

Subgroups of $Z_{36}$ are cyclic groups of any orders which are divisors of 36 (for any divisor such a subgroup is unique). They are: $Z_1=(0)$, $Z_2=(18)$, $Z_3 = (12)$, $Z_4=(9)$, $Z_6=(6)$, $Z_9=(4)$, $Z_{12}=(3)$ , $Z_{18}=(2)$ and $Z_{36}=(1)$ . The subgroup lattice diagram for Z₃₆, expressing the inclusion relation on these subgroups is given below.

$U(12)=\{1,5,7,11\}$ - the group of invertible (by multiplication) elements of $\mathbb{Z}_{12}$ . Since $1^2=5^2=7^2=11^2=1 \mod 12$ , each element has order 2 or 1. The group is isomorphic to $\mathbb{ Z}_2^2$ and its subgroups are $(1), (5), (7), (11), U(12)$ . The subgroup lattice diagram for U(12), expressing the inclusion relation on these subgroups is also given below.