Subgroups and Cyclic Groups

Let $H = 2\mathbb Z = \{\ldots, -4, -2, 0, 2, 4, \ldots\}$ inside $G = (\mathbb Z, +)$. $H$ is nonempty, the sum of two even numbers is even (closed), and the negative of an even number is even (closed under inverses) — so $2\mathbb Z \le \mathbb Z$ by the two-step test. The odd integers fail immediately: $1 + 1 = 2$ is not odd, so they are not even closed, and $0 \notin H$ confirms it is no subgroup.

Is $H = \{0, 4, 8\}$ a subgroup of $\mathbb Z_{12}$ (addition mod $12$)? Since $\mathbb Z_{12}$ is finite, the one-step test applies: check closure alone. $4 + 4 = 8$, $4 + 8 = 12 \equiv 0$, $8 + 8 = 16 \equiv 4$ — every sum lands back in $H$. So $H \le \mathbb Z_{12}$, with no separate check needed for $0$ or for inverses.

Is $H = \{1, 4, 7, 10, 13\}$ a subgroup of $U(15)$? Before testing closure, check that $H \subseteq U(15)$ at all: $U(15)$ consists of the residues coprime to $15$, but $\gcd(10, 15) = 5$, so $10$ is not even a unit mod $15$. $H$ fails to be a subset of $U(15)$, so the question of closure never arises — it cannot be a subgroup.

Is $H = \{1, 4, 11, 14\}$ a subgroup of $U(15)$ under multiplication mod $15$? Each element is coprime to $15$, so $H \subseteq U(15)$. Squaring every element: $4^2 = 16 \equiv 1$, $11^2 = 121 \equiv 1$, $14^2 = 196 \equiv 1$ (all mod $15$) — so every non-identity element is its own inverse. The table below confirms closure: every product of two elements of $H$ lands in $H$. By the two-step test, $H \le U(15)$.

$\langle 1 \rangle = \{n \cdot 1 : n \in \mathbb Z\} = \mathbb Z$, so $1$ generates all of $\mathbb Z$; so does $-1$. But $\langle 2 \rangle = \{\ldots, -4, -2, 0, 2, 4, \ldots\} = 2\mathbb Z$, only the even integers — $2$ generates a proper subgroup, not all of $\mathbb Z$. $\mathbb Z$ is cyclic, but "cyclic" does not mean every element is a generator.

Does $5$ generate $\mathbb Z_{12}$? List the multiples of $5$ mod $12$: $0, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7$, and then back to $0$ — all twelve residues appear before the list repeats, so $\langle 5 \rangle = \mathbb Z_{12}$ and $5$ is a generator. Contrast $4$: its multiples mod $12$ are only $0, 4, 8$, then back to $0$ — $\langle 4 \rangle = \{0, 4, 8\}$, a proper subgroup. The difference is $\gcd(5, 12) = 1$ but $\gcd(4, 12) = 4$; Section 3 explains exactly why the gcd decides this.

$U(21) = \{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\}$ has $12$ elements (the numbers below $21$ coprime to both $3$ and $7$). If $U(21)$ were cyclic, some element would have order $12$. Computing orders directly: $2, 5, 10, 11, 17, 19$ each have order $6$ (for instance $2^6 = 64 = 3\cdot21+1 \equiv 1$, while $2^2 = 4$ and $2^3 = 8$ are not $1$), $4$ and $16$ have order $3$ ($4^3 = 64 \equiv 1$), $8, 13, 20$ have order $2$ ($8^2 = 64 \equiv 1$, $13^2 = 169 \equiv 1$, $20^2 \equiv (-1)^2 = 1$), and $1$ has order $1$. The largest order present is $6$, not $12$ — no element generates the whole group, so $U(21)$ is not cyclic.

In $(\mathbb Z, +)$, does any positive multiple of $1$ ever equal $0$? No — $n \cdot 1 = n \ne 0$ for every positive integer $n$. So $1$ has infinite order, and $\langle 1 \rangle$ is an infinite cyclic subgroup (here, all of $\mathbb Z$). Every element of a finite group has finite order (Section 3), but a cyclic group itself need not be finite.

$\gcd(12, 8) = 4$, so by the order formula $|8| = 12/4 = 3$. Checking directly: $8, 8+8=16\equiv4, 4+8=12\equiv0$ — three steps to return to $0$, confirming the order is $3$.

$\gcd(30, 18) = 6$, so $|18| = 30/6 = 5$. Directly: $18, 36\equiv6, 54\equiv24, 72\equiv12, 90\equiv0$ — exactly five steps, matching the formula.

By the order formula, $|k| = 6$ exactly when $12/\gcd(12,k) = 6$, i.e. $\gcd(12, k) = 2$. Checking $k = 0, \ldots, 11$: only $k = 2$ and $k = 10$ satisfy $\gcd(12, k) = 2$. So $\mathbb Z_{12}$ has exactly $2$ elements of order $6$ — matching the totient $\varphi(6) = 2$, since the generators of any cyclic group of order $6$ (here, the subgroup $\{0,2,4,6,8,10\}$) number $\varphi(6)$.

The order formula assumes $a$ already sits inside a finite cyclic group. In $(\mathbb Z, +)$ itself, $1$ has infinite order (Section 2) — there is no finite $n$ to divide by. The formula only ever applies inside $\langle a \rangle$ once that cyclic subgroup is known to be finite.

The divisors of $12$ are $1, 2, 3, 4, 6, 12$, so $\mathbb Z_{12}$ has exactly $6$ subgroups, one per divisor: order $1$: $\{0\}$; order $2$: $\langle 6\rangle = \{0,6\}$; order $3$: $\langle 4\rangle = \{0,4,8\}$; order $4$: $\langle 3\rangle = \{0,3,6,9\}$; order $6$: $\langle 2\rangle = \{0,2,4,6,8,10\}$; order $12$: $\langle 1\rangle = \mathbb Z_{12}$ itself.

The divisors of $30$ are $1, 2, 3, 5, 6, 10, 15, 30$ — eight of them — so $\mathbb Z_{30}$ has exactly $8$ subgroups, with those eight orders, no computation of any actual elements required.

Find the subgroup of $\mathbb Z_{24}$ of order $4$. Here $n = 24$, $d = 4$, so the formula names the generator at once: $a^{n/d} = 6$. Then $\langle 6 \rangle = \{0, 6, 12, 18\}$, and indeed $6, 12, 18, 24\equiv0$ is four steps, so $|6| = 4$ as required — no guessing was needed to find it.

In $\mathbb Z_{12}$, both $\langle 4 \rangle$ and $\langle 8 \rangle$ equal $\{0, 4, 8\}$: multiples of $4$ give $0,4,8$, and multiples of $8$ give $0,8,4$, the same set in a different order. This is forced by uniqueness: both $4$ and $8$ have $\gcd(12, \cdot) = 4$, so both have order $12/4 = 3$ by the order formula, and a cyclic group of order $12$ has only one subgroup of order $3$ for them to generate.