A Gallery of Groups

Build the addition table for $\mathbb Z_6 = \{0,1,2,3,4,5\}$: row $a$, column $b$ holds $a + b \bmod 6$. For instance row $4$, column $5$ holds $4 + 5 = 9 \equiv 3 \pmod 6$. The table is symmetric across the main diagonal, the Cayley-table signature of an abelian group; every row is also the row above it shifted one step, since adding $1$ slides every entry forward and wraps the overflow back to the front.

Solve $3 + x \equiv 1 \pmod 7$ for $x \in \mathbb Z_7$. Add the inverse of $3$ to both sides: the inverse of $3$ in $(\mathbb Z_7, +)$ is $7 - 3 = 4$, so $x \equiv 1 + 4 = 5 \pmod 7$. Check: $3 + 5 = 8 \equiv 1 \pmod 7$. Unlike solving $3x = 1$ (which needs $3$ to be a unit, the subject of the next section), an additive equation like this one is always solvable, because every element of $(\mathbb Z_n, +)$ has an additive inverse.

List every multiple of $4$ inside $\mathbb Z_{12}$: $4\cdot0 = 0$, $4\cdot1 = 4$, $4\cdot2 = 8$, $4\cdot3 = 12 \equiv 0$ — the list repeats after $3$ steps, giving the set $\{0, 4, 8\}$. This set is closed under addition ($4 + 4 = 8$, $4 + 8 = 12 \equiv 0$, $8 + 8 = 16 \equiv 4$) and contains $0$ and every inverse, so it is a subgroup of $\mathbb Z_{12}$ — the cyclic subgroup generated by $4$, of order $3$ (more on subgroups in Topic 3).

$U(8) = \{1, 3, 5, 7\}$, the four numbers below $8$ coprime to it ($2, 4, 6$ all share the factor $2$). The multiplication table mod $8$ shows every element squares to $1$: $3^2 = 9 \equiv 1$, $5^2 = 25 \equiv 1$, $7^2 = 49 \equiv 1 \pmod 8$. A group in which every non-identity element has order $2$ is the Klein four-group, $U(8) \cong \mathbb Z_2\times\mathbb Z_2$, not the cyclic group of order $4$.

The units mod $15$ are the $8$ numbers below $15$ coprime to both $3$ and $5$: $U(15) = \{1, 2, 4, 7, 8, 11, 13, 14\}$. Computing the order of each by its powers (e.g. $2^1=2, 2^2=4, 2^3=8, 2^4=16\equiv1$, so $\operatorname{ord}(2) = 4$) shows every element has order $1$, $2$ or $4$: only $1$ has order $1$; $4, 11, 14$ have order $2$; and $2, 7, 8, 13$ have order $4$. Since $|U(15)| = 8$ but no element reaches order $8$, $U(15)$ is not cyclic.

In $U(7) = \{1,2,3,4,5,6\}$, the powers of $3$ are $3, 3^2\equiv2, 3^3\equiv6, 3^4\equiv4, 3^5\equiv5, 3^6\equiv1 \pmod7$: all six nonzero residues appear before returning to $1$, so $\operatorname{ord}(3) = 6 = |U(7)|$, and $U(7) = \langle3\rangle$ is cyclic. In fact $U(n)$ is cyclic exactly when $n$ is $1$, $2$, $4$, $p^k$, or $2p^k$ for an odd prime $p$ — which is why $U(7)$ succeeds and $U(15) = U(3\cdot5)$ above does not.

The multiplication table of $D_3 = \{e, r, r^2, s, rs, r^2s\}$ (symmetries of an equilateral triangle) is NOT symmetric across the diagonal, unlike every table drawn so far: the entry in row $s$, column $r$ is $sr = r^2s$, while the entry in row $r$, column $s$ is $rs$ — a different element. This asymmetry is the Cayley-table signature of a non-abelian group.

Label the triangle's vertices $0,1,2$ counterclockwise; let $r$ rotate $0\to1\to2\to0$ and let $s$ fix vertex $0$ while swapping $1,2$. Apply $r$ then $s$ (that is, $sr$): $0\xrightarrow r1\xrightarrow s2$, $1\xrightarrow r2\xrightarrow s1$, $2\xrightarrow r0\xrightarrow s0$, so $sr$ sends $0\to2, 1\to1, 2\to0$ — the reflection $r^2s$, fixing vertex $1$. Apply $s$ then $r$ (that is, $rs$): $0\xrightarrow s0\xrightarrow r1$, $1\xrightarrow s2\xrightarrow r0$, $2\xrightarrow s1\xrightarrow r2$, so $rs$ sends $0\to1, 1\to0, 2\to2$ — the reflection $rs$, fixing vertex $2$ instead. Since $sr = r^2s$ fixes vertex $1$ while $rs$ fixes vertex $2$, they are different reflections: $sr \ne rs$.

$D_6$, the symmetries of a regular hexagon, has order $12 = 2\cdot6$: six rotations $e, r, r^2, r^3, r^4, r^5$ and six reflections $s, rs, r^2s, \ldots, r^5s$. Every reflection has order $2$, since reflecting twice returns every vertex to its start — that is already $6$ elements. Among the rotations, $r^k$ has order $2$ only when $r^{2k} = e$, i.e. $6 \mid 2k$, i.e. $3\mid k$: only $k = 3$ works among $0,\ldots,5$, giving the single rotation $r^3$ (the $180°$ rotation). So $D_6$ has exactly $6 + 1 = 7$ elements of order $2$: the $6$ reflections together with $r^3$.

Let $A = \begin{pmatrix}1&2\\0&1\end{pmatrix}$ and $B = \begin{pmatrix}1&0\\3&1\end{pmatrix}$; both have determinant $1$, so both lie in $GL(2,\mathbb R)$. Multiplying in one order, $AB = \begin{pmatrix}7&2\\3&1\end{pmatrix}$; in the other order, $BA = \begin{pmatrix}1&2\\3&7\end{pmatrix}$. Since $AB \ne BA$, $GL(2,\mathbb R)$ is non-abelian — confirmed by nothing more exotic than $2\times2$ matrix multiplication.

Find the inverse of $M = \begin{pmatrix}2&3\\1&2\end{pmatrix}$. First $\det M = 2\cdot2 - 3\cdot1 = 1 \ne 0$, so $M$ is invertible. The $2\times2$ inverse formula swaps the diagonal, negates the off-diagonal, and divides by the determinant: $M^{-1} = \dfrac1{\det M}\begin{pmatrix}2&-3\\-1&2\end{pmatrix} = \begin{pmatrix}2&-3\\-1&2\end{pmatrix}$ (since $\det M = 1$). Check: $MM^{-1} = \begin{pmatrix}2&3\\1&2\end{pmatrix}\begin{pmatrix}2&-3\\-1&2\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I$.

Over the field $\mathbb Z_2 = \{0,1\}$ there are $2^4 = 16$ matrices $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ in all; a matrix is invertible exactly when $\det = ad - bc \not\equiv 0 \pmod2$. Checking all $16$ leaves exactly $6$ invertible matrices, so $|GL(2,\mathbb Z_2)| = 6$. Multiplying them out shows the group is non-abelian with one identity, three elements of order $2$, and two of order $3$ — exactly the order pattern of $D_3$ from the previous section: $GL(2,\mathbb Z_2) \cong D_3$, the smallest non-abelian group, realized here as matrices instead of triangle symmetries.

Find $\operatorname{ord}(8)$ in $\mathbb Z_{20}$. Add $8$ to itself repeatedly: $8, 16, 24\equiv4, 12, 20\equiv0$ — reached $0$ after $5$ steps, with no earlier partial sum equal to $0$. So $\operatorname{ord}(8) = 5$ in $\mathbb Z_{20}$, matching the general formula $\operatorname{ord}(a) = n/\gcd(n,a)$ for $(\mathbb Z_n,+)$: here $20/\gcd(20,8) = 20/4 = 5$.

Find $\operatorname{ord}(2)$ in $U(11)$. The powers of $2$ mod $11$ are $2, 4, 8, 16\equiv5, 10, 20\equiv9, 18\equiv7, 14\equiv3, 6, 12\equiv1$ — all ten nonzero residues appear before returning to $1$, so $\operatorname{ord}(2) = 10 = |U(11)|$: $2$ is a generator of $U(11)$.

By the formula $\operatorname{ord}(g^k) = \dfrac{\operatorname{ord}(g)}{\gcd(\operatorname{ord}(g),k)}$ (proved below), $\operatorname{ord}(g^9) = \dfrac{12}{\gcd(12,9)} = \dfrac{12}3 = 4$. Test it concretely with $g = r$, the rotation of order $12$ in $D_{12}$: $r^9$ applied four times is $r^{36} = r^{36\bmod12} = r^0 = e$, and $r^9, r^{18} = r^6, r^{27} = r^3$ are all different from $e$, so $4$ really is smallest.