Generated Subgroups, Products, and Isomorphisms

December 1, 2025 · Luciano Muratore

It is well-known that groups can be built up from smaller pieces. A natural question is: given a subset $S$ of a group $G$, what is the smallest subgroup of $G$ that contains $S$?

Generated Subgroups

Let $(G, \cdot)$ be a group and $S \subseteq G$. The subgroup generated by $S$, denoted $\langle S \rangle$, is defined as the smallest subgroup of $G$ containing $S$, or equivalently,

$\langle S \rangle = \bigcap \{H \leq G \mid S \subseteq H\}.$

Intuitively: start with $S$ and close under products and inverses until nothing more can be added without leaving a subgroup. But what does $\langle S \rangle$ look like concretely?

The answer is that every element of $\langle S \rangle$ is a finite product of elements of $S$ and their inverses. More precisely, setting $H := \{1, s_1 s_2 \cdots s_n \mid s_i \in S \cup S^{-1}\}$, one can check that $H$ is a subgroup: $1 \in H$ by the empty product, a product of two words in $S \cup S^{-1}$ is again a word in $S \cup S^{-1}$, and the inverse of $s_1 \cdots s_n$ is $s_n^{-1} \cdots s_1^{-1}$, which is also in $H$. Since $H$ contains $S$ and any subgroup containing $S$ must contain all finite products of its elements and their inverses, we conclude that $H = \langle S \rangle$.

The Dihedral Group

A beautiful example of a generated subgroup is the dihedral group $D_n$. Consider a regular convex $n$-gon in the plane. The dihedral group $D_n$ is the group of all isometries of the plane that map the $n$-gon to itself, with composition as the operation. It has $|D_n| = 2n$ elements: $n$ rotations about the center and $n$ reflections with respect to lines through the center.

Label the vertices $1, 2, \ldots, n$ in cyclic order, let $r$ be the counterclockwise rotation by $\frac{2\pi}{n}$, and let $s$ be a reflection through, say, vertex $1$ and the center. Then $D_n = \langle r, s \rangle$.

Why? The rotations are $\{1, r, r^2, \ldots, r^{n-1}\} = \langle r \rangle$. For the reflections, consider the conjugates $r^j s r^{-j}$ for $j \in \{0, \ldots, n-1\}$. Geometrically, $r^{-j}$ rotates the polygon, $s$ reflects across a fixed axis, and $r^j$ rotates back — the net effect is a reflection across the axis obtained by rotating the original axis by $j \cdot \frac{2\pi}{n}$. So every reflection appears as such a conjugate, and since $D_n$ consists only of rotations and reflections, we get $D_n = \langle r, s \rangle$.

Groups of Exponent 2 are Abelian

Here is a small but elegant observation. Let $(G, \cdot)$ be a group such that $g^2 = 1$ for all $g \in G$ — that is, every element has order dividing 2. Then $G$ is abelian.

The key observation is that $g^2 = 1$ implies $g = g^{-1}$ for all $g \in G$. Take any $a, b \in G$. By hypothesis, $(ab)^2 = 1$, so $(ab)^{-1} = ab$. But also $(ab)^{-1} = b^{-1}a^{-1} = ba$, since every element is its own inverse. Therefore $ab = ba$, and since this holds for all $a, b \in G$, the group is abelian.

Direct Products

Given two groups $(G, \ast)$ and $(H, \star)$, their direct product $G \times H$ is the set of pairs $\{(g, h) \mid g \in G, h \in H\}$ with componentwise operation:

$(g_1, h_1) \cdot (g_2, h_2) = (g_1 \ast g_2, h_1 \star h_2).$

The intuition is simple: operate in $G$ on the first component and in $H$ on the second, independently. One can verify that this makes $(G \times H, \cdot)$ a group: closure and associativity follow from those in $G$ and $H$; the identity is $(e_G, e_H)$; and the inverse of $(g, h)$ is $(g^{-1}, h^{-1})$.

A concrete and important example is $\mathbb{Z}_2 \times \mathbb{Z}_2$, the Klein group $V_4$. It has four elements $\{(0,0),(1,0),(0,1),(1,1)\}$, every non-identity element has order 2, and the group is abelian. Interestingly, $V_4$ appears not only as an abstract product but also inside $S_4$: the set $\{e, (12)(34), (13)(24), (14)(23)\}$ — the identity and the three bitranspositions — forms a subgroup of $S_4$ isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$.

Some Fundamental Isomorphisms

Let me illustrate how direct products and isomorphisms interact through a few examples.

$(\mathbb{R}^\times, \cdot) \cong (\mathbb{R}^+, \cdot) \times (\mathbb{Z}_2, +)$. Every nonzero real can be written as $x = \text{sgn}(x) \cdot |x|$, and encoding the sign in $\mathbb{Z}_2$ gives a bijective homomorphism $\varphi(x) = (|x|, 0)$ if $x > 0$ and $(|x|, 1)$ if $x < 0$.

$(\mathbb{Z}_4, +) \cong (\{1,-1,i,-i\}, \cdot)$. The map $\psi([k]_4) = i^k$ is a homomorphism since $\psi(a+b) = i^{a+b} = i^a \cdot i^b = \psi(a)\psi(b)$, and it is bijective since $i^0, i^1, i^2, i^3$ are distinct and exhaust the target.

$(\mathbb{Z}_6, +) \cong (\mathbb{Z}_2 \times \mathbb{Z}_3, +)$. Define $\theta([k]_6) = ([k]_2, [k]_3)$. This is a homomorphism by the compatibility of reduction modulo 2 and 3. It is injective because $\theta([k]_6) = (0,0)$ forces $k$ to be divisible by both 2 and 3, hence by 6. Since both groups have order 6, injectivity implies bijectivity.

The last example is a special case of a more general fact: $\mathbb{Z}_{mn} \cong \mathbb{Z}_m \times \mathbb{Z}_n$ whenever $\gcd(m, n) = 1$. I would let the readers convince themselves of this.

Translations of a Vector Space

Let me close with a charming observation about translations. Let $V$ be a vector space over a field $K$. For any $v \in V$, the translation by $v$ is the map $\tau_v : V \to V$ defined by $\tau_v(x) = x + v$. The set of all translations $T = \{\tau_v \mid v \in V\}$, under composition, forms a group: $\tau_v \circ \tau_w = \tau_{v+w}$, the identity is $\tau_0$, and the inverse of $\tau_v$ is $\tau_{-v}$.

The map $\Phi : V \to T$ defined by $\Phi(v) = \tau_v$ is an isomorphism. It is a homomorphism since $\Phi(v + w) = \tau_{v+w} = \tau_v \circ \tau_w = \Phi(v) \circ \Phi(w)$. It is injective because $\Phi(v) = \Phi(w)$ implies $x + v = x + w$ for all $x$, hence $v = w$. And it is surjective by definition of $T$.

So $(V, +) \cong (T, \circ)$: the additive group of a vector space is nothing more than the group of its own translations. The abstract and the geometric are the same thing, dressed differently.