Generating and presenting groups

Section 1.4 Generating and presenting groups

Long products are group elements are important to us. The following lemma gives us a definition of a subgroup generated by a collection of elements

Lemma 1.4.1.

Let \(H \leq G\) be a subgroup and let \(S=\{h_1,\ldots\} \subset G\) be a set of elements. The following are equivalent:

\(H\) is the smallest subgroup of \(G\) containing \(S\text{,}\) i.e.
\begin{equation*} H = \bigcap_{S\subset K \leq G} K, \end{equation*}
i.e. the intersection of all subgroups of \(G\) containing \(S\text{.}\)
\(H\) consists of all products of elements \(S^{\pm 1}\text{,}\) i.e.
\begin{equation*} H = \left\{h_{i_1}^{\epsilon_1}\cdots h_{i_n}^{\epsilon_n} \in G \mid h_{i_j}\in S, \epsilon_j \in \{\pm 1\}, n \in \ZZ_{\geq 0} \right\} \end{equation*}

We will denote the subgroup generated by \(S=\{h_1,\ldots\}\) as \(\gen S\) or \(\gen{h_1,\ldots}\text{.}\)

Proof.

Let us denote by \(H\) the smallest subgroup containing \(S\) and by \(\gen S\) set of products of elements in \(S^{\pm 1}\text{.}\) \(\gen S\) is closed under multiplication and taking inverses so it's a subgroup of \(G\text{.}\) By definition, it's therefore immediate that \(H \subset \gen S\text{.}\)

To see that \(H \supset \gen S\text{,}\) we proceed by induction on the length of a product. By definition any product of length 1, being an element of \(S^{\pm 1}\text{,}\) is in \(H\text{.}\) If every product of length at most \(n\) is in \(H\text{,}\) then, because a product of length \(n+1\) is a product of a product of length \(n\) and a product of length \(1\) and because \(H\text{,}\) being a subgroup, is closed under multiplication, every product of length \(n+1\) is also in \(H\text{.}\) Therefore, by induction, \(H \supset \gen S\) and equality follows.

Any set \(S \subset G\) such that \(G = \gen S\) is called a generating set of G. If \(G\) has a finite generating subset \(\{g_1,\cdots,g_k\}\subset G\) then we way \(G\) is finitely generated. The rank of a group \(G\) is the minimal cardinality achieved by its generating sets, though this terminology is less standard and typically has other meanings in different group theoretical contexts.

Since \(G = \gen G\text{,}\) every finite group is finitely generated and it is a standard fact that every symmetric group \(S_n\) has rank 2. It's already more impressive when an infinite group has finite rank. Here are three examples:

If \(X = \{a,b,c\}\) then the free group \(F(X)\) has rank 3. In general if \(|X|=n\) then we will say that \(F(X)\) is a free group of rank \(n\text{.}\)
\(\GL 2 \ZZ\text{,}\) the group of \(2\times 2\) invertible matrices with \(\ZZ\) coefficients also has rank 2, indeed it is generated by elementary matrices:
\begin{equation*} \GL 2 \ZZ = \gen{ \begin{bmatrix} 0&1\\ 1&0 \end{bmatrix}, \begin{bmatrix} 1&1\\ 0&1 \end{bmatrix} }. \end{equation*}
\(\pi_1(M)\text{,}\) where \(M\) is a compact finite dimensional manifold is also finitely generated.

If \(\gen S =G\text{,}\) then by Theorem 1.3.1, there is an epimorphism, or a surjective homomorphism, \(\varphi:F(S) \onto G\text{.}\) So it is immediate that if \(G\) has rank \(n\) then it is the homomorphic image of a free group of rank \(n\text{.}\)

In any case if \(G = \gen S\) and \(X\) is any alphabet such that \(|X| = |S|\text{,}\) by the First Isomorphism Theorem for groups we have

\begin{equation*} F(X)/\ker(\varphi) \approx G. \end{equation*}

Let us fix some more terminology before continuing. If \(g,h \in G\text{,}\) then the conjugate of \(g\) by \(\)h is

\begin{equation*} g^h = h^{-1} g h. \end{equation*}

Some people write \(g^h = h g h^{-1}\text{,}\) but they are wrong because we should have \((g^h)^k = g^{hk}\text{.}\) Although it looks weird, it makes more sense to write

\begin{equation*} {~}^h g = h gh^{-1} = g^{h^{-1}}. \end{equation*}

The correct notation is consistent with the concepts of left and right actions.

Recall that a normal subgroup \(N \triangleleft G\) is a subgroup that is closed under conjugation by arbitrary elements in \(G\text{,}\) and the class of normal subgroups coincides with the kernels of homomorphisms from \(G\text{.}\) Consider now the concept of normal generation:

Proposition 1.4.2.

Let \(R\subset G\) be some subset. Then the smallest normal subgroup of \(G\) containing \(R\) coincides with

\begin{equation*} \ncl R = \left\{\prod_{i=1}^n c_i^{-1}r_i^{\epsilon_i} c_i \middle | r_i \in R,\epsilon_i \in \{\pm 1\}, c_i \in G, n \in \ZZ_{\geq 0}\right\}, \end{equation*}

the set of products of conjugates of elements in \(R^{\pm 1}\text{.}\)

We leave the proof to Exercise 1.4.1.1.

Consider a subset \(R \subset F(X)\) is a subset then we will call a pair \(\pres X R\) a group presentation. The set \(X\) of letters is called the generators and words in \(R\) are called relations.

Again we will consider \(\pres X R\) to be a set of words in \(X^{\pm 1}\) subject to the following equality relation: \(w =_{\pres X R} w'\) if there is a sequence of words:

\begin{equation*} w = w_0 \to w_1 \to \cdots \to w_m = w' \end{equation*}

such that each \(w_i \to w_{i+1}\) is either

An elementary insertion or cancellation given in Definition 1.2.1,
the deletion of a subword that is a relation, i.e.
\begin{equation*} w_{i} = w'rw'' \to w'w'' = w_{i+1}, \end{equation*}
for some \(r \in R\text{,}\) or
the insertion of some relation into a word, i.e.
\begin{equation*} w_{i} = w'w'' \to w'rw'' = w_{i+1}, \end{equation*}
for some \(r \in R\text{.}\)

So again, elements of \(\pres X R\) are equivalence classes of words. Once again we want multiplication to be given by concatenation. Recall that for free groups we showed that concatenation gave a well defined product on the set of equivalence classes of words related by sequences of elementary insertions or deletions. Our argument relied on reduced words. In this case, we don't have reduced words so instead we will condinue to extend the relation \(=_{\pres X Y}\) as follows: if \(w=_{\pres X R}w'\) and \(u=_{\pres X R}u'\) then we impose \(w*u =_{\pres X R} w'*u'\text{.}\) We can therefore continue to extend \(=_{\pres X R}\) in this way until we make \(\pres X R\) into a group. Now maybe this group collapsed down to the trivial group, but it's a group.

So on the one hand we have that every group is a quotient of a free group, on the other hand we have constructed the group \(\pres X R\) and we have.

Theorem 1.4.3.

\begin{equation*} \pres X R \approx F(X)/\ncl R \end{equation*}

If \(G \approx \pres X R\) with \(|X| \leq \infty\text{,}\) then we say \(G\) is finitely generated. If both \(|X|,|R|\) are finite, the we say \(G\) is finitely presented.

Exercises 1.4.1 Exercises

1.

Prove Proposition 1.4.2.

Hint: This is similar to the proof of Lemma 1.4.1 the only issue is that if you take a conjugate of a product of conjugates, i.e.

\begin{equation*} g^{-1} \left(\prod_{i=1}^n \left(c_i^{-1}r_i c_i\right) \right)g, \end{equation*}

then it's not clear that the result is again a product of conjugates. See what happens when you conjugate each factor of the product by \(g\text{.}\)

2.

Let \(w \in F(X)\) be written as a word, let \(R\) be a subset of \(F(X)\) and consider the new word \(w'\) obtained by inserting some element \(r \in R\text{,}\) i.e.

\begin{equation*} w = w_1*w_2 \to w_1*r*w_2. \end{equation*}

Show that there are elements \(h,k \in \ncl R\) such that:

\begin{equation*} hw = w' = wk. \end{equation*}

Hint: conjugate \(r\) by a prefix or suffix of of \(w\text{.}\)

3.

Prove Theorem 1.4.3.

Hint: Take \(\varphi: F(X) \onto \pres X R\text{.}\) Use the universal property of free groups as well as the previous exercise and argue that \(\ker{\phi} = \ncl R.\)

4.

Use Proposition 1.4.2 and Theorem 1.4.3 to argue that equality in \(\pres X R\) is fully defined by elementary insertion and deletions and insertions and removals of relations as subwords. Forcing products to be well defined is superfluous.

Hint: By Theorem 1.4.3, \(w=_{\pres X R} w'\) if and only if there is some \(n \in \ncl R\) such that \(wn = w'\text{.}\)