# 3.2: Subgroups

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

When we consider the symmetries of, say, a pentagon, we notice that it has rotational symmetries like the 'bumpy' pentagon. From the bumpy pentagon, we see that the rotations themselves form a group; there's a group of rotations inside the group of symmetries of the pentagon! Likewise, if we consider just the flip, we see a group similar to the symmetries of the perfectly symmetrical face. This yields another group inside the symmetries of the pentagon. We can make this precise:

Definition 3.2.0: Subgroup

Let \(G\) be a group, and \(H\) a subset of \(G\). Then \(H\) is a *subgroup* of \(G\) if \(H\) is itself a group using the same operation as \(G\).

Ostensibly, to check that a subset \(H\) is a subgroup, we would need to check all four properties of the group. That is, closure (ie, the operation gives a map \(H\times H\rightarrow H\); products of things in \(H\) are always in \(H\)), identity, the existence of inverses, and associativity.

In fact, since \(H\) has the same operation as \(G\), we know that the operation in \(H\) is associative (since \(G\) is a group). Furthermore, if the operation is closed and inverses exist, then we know that for any \(h\in H\), \(hh^{-1}=e\) must be in \(H\). So really we only need to check two things:

- Closure: \(gh\in H\) for all \(g,h\in H\), and
- Inverses: \(h^{-1}\in H\) for all \(h\in H\).

Some important things to notice:

- The group \(G\) is always a subgroup of itself! (\(G\) is a subset of itself, which is a group with the same operation as \(G\).)
- The subset containing just the identity element is also a subgroup! This is called the trivial subgroup.
- The set of all powers of an element \(h\) (\(\{\ldots, h^{-1}, h^{-2}, e, h, h^2, \ldots\}\)) is a subgroup of \(G\). This is called the cyclic subgroup generated by \(h\).

Exercise 3.2.1

Let \(X\) be a geometric object. Show that the rotations of \(X\) back onto itself forms a subgroup of the group of symmetries of \(X\). (Try this in particular on a regular polygon and a regular polyhedron. What happens with a 'bumpy' polygon?)

Let \(G\) be a group, and \(g\in G\). Consider a function \(f_g:G\rightarrow G\) given by \(f_g(h)=g\cdot h\). (This is the 'left multiplication by \(g\)' function.) What happens if, for some \(h, k \in G\), \(f_g(h)=f_g(k)\)? Then \(gh=gk\), so \(g^{-1}gh=g^{-1}gk\), and \(h=k\). This tells us that \(f_g\) is a one-to-one, or injective, function. If \(G\) has a finite number of elements, then \(f_g\) is also an onto function, and is thus a bijection from \(G\) back to itself. Then we can consider \(f_g\) as a permutation of \(G\)!

If we consider \(G\) as a set, we can think of any left multiplication as a permutation of \(G\). But the set of all left multiplications is itself a group. This gives us what is known as Cayley's Theorem!

Theorem 3.2.2: Cayley's Theorem

The ideal gas law is easy to remember and apply in solving problems, as long as you get the **proper values a**

Exercise 3.2.3

Label the six symmetries of the equilateral triangle. Demonstrate that the symmetries of the triangle are a subgroup of \(S_6\), the permutations of \(6\) objects.

It is worth noticing that for any \(g\) in a group \(G\), the powers of \(g\) generate a subgroup of \(G\). The set \(\{g^i \mid i \in \mathbb{Z} \}\) is closed under the group operation, and includes the identity and inverses. This is called the cyclic subgroup generated by \(g\).

Exercise 3,2,4

Find all of the subgroups of the permutation group \(S_3\) for three objects. Which subgroups are subgroups of other subgroups? Name each subgroup, and arrange them according to which is contained in which.

### Contributors

- Tom Denton (Fields Institute/York University in Toronto)