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For any cyclic subgroup $H=⟨s⟩$ of order $n\in \{1,2,4\}$, defineBoth preserve products because scalar phases and planar rotations add their angles. Each sends the generator to an element of exact order $n$, so both have trivial kernel and are faithful representations.

For $⟨U⟩\cong C_4$, take $s=U$: the generator images are $i$ and $R(\pi /2)$. For each order-two subgroup take its nonidentity element $s$; the images are $-1$ and $-I_2$. For the trivial subgroup the unique images are $1$ and $I_2$, which are faithful on a one-element group.

The one-dimensional representation of $C_4$ here must use complex scalars. There is no faithful one-dimensional real representation of $C_4$, since a nonzero real scalar of finite order is $1$ or $-1$, and neither has order four. The two-dimensional representations above are real.

For a normal subgroup $H$, the quotient group consists of its cosets with multiplicationTo check that this is well-defined, replace $g,k$ by $g{h}_1,k{h}_2$ with $h_1,h_2\in H$. Their product issince normality puts $k^{-1}{h}_{1}k$ in $H$. Hence the product does not depend on representatives. Associativity follows from that in $G$, the identity element is $H=eH$, and $(gH{)}^{-1}=g^{-1}H$. These verify the group axioms.

Lagrange's theorem counts the equal-sized cosets:Finally, if $|G|=p$ is prime, choose $g\ne e$. Its cyclic subgroup has order greater than one dividing $p$, hence equals $G$. All elements are powers of $g$, so $g^j{g}^k=g^{j+k}=g^k{g}^j$. Thus

Act on the four labelled vertices. A proper rotation gives an even permutation, identifying the rotational tetrahedral symmetry group with $A_4$, an alternating group. Geometrically there are one identity, three half-turns about opposite-edge midpoint axes, and two nontrivial rotations about each of four vertex-face axes. ThereforeThe tetrahedral rotation conjugacy classes areTo justify the splitting, the centralizer of a 3-cycle within $A_4$ is its three-element cyclic subgroup, so its conjugacy class has size $12/3=4$. Its inverse is not conjugate to it in $A_4$: the permutations that reverse that cycle while preserving its fixed vertex are odd. Conjugating $(123)$ by the even permutations produces the displayed $C_{+}$, and taking inverses gives $C_{-}$. The three half-turns are conjugate, since rotations move their axes into one another. Their centralizer has order four and their class therefore has size three.

Distinguish the quaternion group generators $\mathbf{i},\mathbf{j},\mathbf{k}$ from the complex scalar $i$. Denote their proposed image matrices byDirect multiplication givesFor example $AB=\mathrm{diag}(i,-i)=C$ and $ABC=C^2=-I_2$. Sending $I$ to $I_2$ and each negative quaternion element to the negative of its positive image now reproduces every quaternion product. The matrices all have determinant one and are invertible, so this is a group representation in $GL(2,\mathbb{C})$.

The eight matrices $\pm I_2,\pm A,\pm B,\pm C$ are distinct. Therefore the group homomorphism is injective, and the representation is faithful; equivalently its kernel is $\{I\}$.

The character of a representation is the trace of its representing matrix. ConsequentlyThese are constant on the five conjugacy classes $\{I\}$, $\{-I\}$, $\{\pm \mathbf{i}\}$, $\{\pm \mathbf{j}\}$, $\{\pm \mathbf{k}\}$.

The complete list of subgroups isTo see completeness, Lagrange's theorem allows only orders $1,2,4,8$. The only element of order two is $-I$, which fixes the order-two subgroup. Every noncentral element has order four and generates the corresponding listed cyclic subgroup. Any order-four subgroup contains such an element, because there are only two central elements; it must equal the cyclic subgroup generated by that element. This accounts for all possibilities.

Let $d=\mathrm{ord}(g)$. The powersare distinct: equality of two with exponents between $0$ and $d-1$ would give a smaller positive power equal to $e$. Reducing exponents modulo $d$ proves closure and supplies inverses, so this is a cyclic subgroup of group order $d$. Lagrange's theorem now gives

Choose $g\ne e$ in a group of prime group order $p$. Its order of a group element divides $p$ by (c) and is not $1$, hence equals $p$. Consequently its cyclic subgroup already contains all elements of $G$:Thus every nonidentity element is a group generator.

Let commuting elements $a,b$ have finite coprime orders $m,n$. Their cyclic subgroups have trivial intersection by Lagrange's theorem. If $(ab{)}^k=1$, then $a^k=b^{-k}$ belongs to this intersection, so $m\mid k$ and $n\mid k$. Thus $mn\mid k$. Conversely, $(ab{)}^{mn}=1$ because the elements commute. This proves the displayed order formula.

A subgroup $H\le G$ is a subset of a group $G$ that is itself a group under the same group operation. It must contain the identity element of $G$ and be closed under products and inverse elements. Associativity is inherited from $G$, so it needs no new proof. The whole group and the trivial group $\{e\}$ are always subgroups. A proper subgroup satisfies $H\ne G$.

A useful single-condition test isTo see why it works, choose $a\in H$; then $a{a}^{-1}=e\in H$. Taking $a=e$ gives every inverse element $b^{-1}\in H$. Applying the test to $a$ and $b^{-1}$ then gives $ab\in H$. Conversely, every subgroup satisfies the test by closure under products and inverse elements. This also makes the additive version memorable: a nonempty subset of an additive group is a subgroup exactly when it is closed under differences. See McKernan's MIT lecture 2, Proposition 2.3, for the equivalent closure criterion.

For the square dihedral group $D_4$ above, $R=\{e,r,r^2,r^3\}$ is a cyclic subgroup: products add exponents modulo four, and $(r^k{)}^{-1}=r^{-k}$. The smaller collections $\{e,r^2\}$ and $\{e,s\}$ are subgroups of group order two. But $\{e,r\}$ is not a subgroup, since $r\cdot r=r^2$ lies outside it. A plausible-looking list of symmetries therefore need not constitute a complete collection of reversible operations.

Inside $(\mathbb{Z},+)$, the even numbers $2\mathbb{Z}$ form a subgroup: $2m-2n=2(m-n)$ is an even number. Its two cosets consist of the even numbers and the odd numbers, so its index of a subgroup is two. The parity group homomorphism $k\mapsto k\mathrm{mod}2$ has $2\mathbb{Z}$ as its kernel of a group homomorphism. A subgroup can thus identify exactly which information a composition-preserving map forgets.

Subgroups also describe residual symmetry. In the group action of $D_4$ on the four vertices, the stabilizer subgroup of $(1,1)$ is $\{e,sr\}$: the diagonal reflection $sr$ fixes that vertex, and the other six motions move it. Requiring a feature to stay fixed still leaves a group, because both composition and reversal preserve that requirement. This is why subgroups are useful for studying a group action one feature at a time.

Fix a real angle fraction $\alpha$, and repeatedly multiply by $g=e^{2\pi i\alpha }$ in the circle group. The generated cyclic subgroup is $H_{\alpha }=\{g^k:k\in \mathbb{Z}\}$. Equivalently, the angle fractions are the fractional parts of $k\alpha$, with one full turn identified with zero.

If $\alpha =p/q$ is a rational number in lowest terms, with $q>0$ and $p,q$ coprime, then $g^q=1$ and no smaller positive power is $1$. Hence $H_{\alpha }$ is a finite cyclic group of order $q$, consisting of equally spaced points. For $\alpha =2/7$, the successive angle fractions, starting at $k=0$, areThe subgroup is finite, so it is a closed subgroup of the circle group; it is also a discrete subgroup. It is not an open subgroup, since every ambient neighbourhood of any of its points contains an arc, rather than only these seven points.

If $\alpha$ is an irrational number, no nonzero integer $k$ satisfies $k\alpha \in \mathbb{Z}$, so the powers never repeat. The stronger statement is that $H_{\alpha }$ is a dense set in the circle group: every nonempty open arc contains a power $g^k$. Thus its closure is all of $\mathbb{T}$, while $H_{\alpha }$ itself is a countable proper subgroup and is not a closed subgroup or a discrete subgroup. See irrational rotation and everywhere interval frequency under an irrational rotation for the corresponding motion and its long-term frequencies.

For $\alpha =\sqrt{2}-1$, direct computation of the fractional parts of $k\alpha$ gives, to six decimal places,For example, $\{3\alpha \}=3\sqrt{2}-4\approx 0.242641$ and $\{5\alpha \}=5\sqrt{2}-7\approx 0.071068$. These calculations specify exactly which points the diagram displays.

This example shows why algebraic generation and closure must be distinguished: one irrational rotation generates a countable cyclic subgroup, yet its closure is the whole uncountable compact group.