Tag Archives: group presentation

Prove that a group with a given presentation is infinite

Prove that G = \langle x,y \ |\ x^3 = y^3 = (xy)^3 = 1 \rangle is an infinite group as follows. Let p be a prime congruent to 1 mod 3 and let G_p denote the nonabelian group of order 3p. Let a,b \in G_p with |a| = p and |b| = 3. Prove that ab and ab^2 have order 3. Deduce that G_p is a homomorphic image of G, and from this deduce that G is infinite, using the fact that there are infinitely many primes congruent to 1 mod 3. [Note that every nonidentity element of G_p has order 3 or p.]


Let p be a prime congruent to 1 mod 3, and let G_p = Z_p \rtimes_\varphi Z_3, where Z_p = \langle a \rangle, Z_3 = \langle b \rangle, and \varphi(b)(a) = \gamma(a) for some order 3 automorphism \gamma \in \mathsf{Aut}(Z_p).

In particular, \gamma(a) = a^k for some k \not\equiv 1 mod p, and k^3 \equiv 1 mod p. Note that k^2 + k + 1 = (k^3-1)/(k-1) \equiv 0/(k-1) \equiv 0 mod p.

Now we compute (a,b)^3: (a,b)(a,b)(a,b) = (a \varphi(b)(a), b^2)(a,b) = (a \varphi(b)(a) \varphi(b^2)(a), b^3) = (a \cdot a^k \cdot a^{k^2}, b^3) = (a^{1+k+k^2},b^3) = (1,1). Thus |(a,b)| = 3.

Similarly, it is easy to see that |(a,b^2)| = 3.

Now let S = \{x,y\} and \theta : S \rightarrow G_p be defined by \theta(x) = ab, \theta(y) = b. Clearly ab and b generate G_p. By the universal property of free groups there exists a unique group homomorphism \Psi : F(S) \rightarrow G_p extending \theta; moreover, because (as computed above) ab and b satisfy b^3 = (ab)^3 = (abb)^3 = 1, there exists a unique group homomorphism \Phi : G \rightarrow G_p such that \Phi \circ pi = \Psi. Since \Psi is surjective, so is \Phi. Thus G_p is a homomorphic image of G.

Since there are infinitely many primes congruent to 1 mod 3, G cannot have finite order.

An automorphism of Sym(6) that is not inner

This exercise exhibits an automorphism of S_6 that is not inner (hence it shows that [\mathsf{Aut}(S_6) : \mathsf{Inn}(S_6)] = 2. Let u_1 = (1\ 2)(3\ 4)(5\ 6), u_2 = (1\ 4)(2\ 5)(3\ 6), u_3 = (1\ 3)(2\ 4)(5\ 6), u_4 = (1\ 2)(3\ 6)(4\ 5), and u_5 = (1\ 4)(2\ 3)(5\ 6). Show that u_1, \ldots, u_5 satisfy the following relations:

  • u_i^2 = 1 for all i,
  • (u_iu_j)^2 = 1 for all i,j with |i-j| \geq 2, and
  • (u_iu_{i+1})^3 = 1 for all i \in \{1,2,3,4\}.

Deduce that S_6 = \langle u_1, \ldots, u_5 \rangle and that the map (1\ 2) \mapsto u_1, (2\ 3) \mapsto u_2, (3\ 4) \mapsto u_3, (4\ 5) \mapsto u_4, (5\ 6) \mapsto u_5 extends to an automorphism of S_6 (which is clearly not inner since it does not preserve cycle shape.


Showing that u_i satisfy the given relations is a straightforward computation. Moreover, they generate S_6 since u_3u_5u_1 = (6\ 5) and u_1u_2u_4 = (6\ 5\ 2\ 3\ 4\ 1).

Count the number of automorphisms of a group from a presentation

Use presentations to find the orders of the automorphism groups of Z_2 \times Z_4 and Z_4 \times Z_4.


We have Z_2 \times Z_4 = \langle a,b \ |\ a^2 = b^4 = 1, ab = ba \rangle.

Now any element x of order 4 and any element y of order 2 with x \neq y^2 generate Z_2 \times Z_4. There are 4 elements of order 4: b, b^3, ab, and ab^3. There are 2 elements of order 2: a, b^2, and ab^2. Once x is chosen, there are two choices for y, and each choice determines an automorphism. These are distinct by construction. Thus |\mathsf{Aut}(Z_2 \times Z_4)| = 8.

Now Z_4 \times Z_4 = \langle a,b \ |\ a^4 = b^4 = 1, ab = ba \rangle.

Any two elements of order 4, say x and y, generate Z_4 \times Z_4 provided \langle x \rangle \cap \langle y \rangle = 1. This group has 12 elements of order 4, and these intersect nontrivially pairwise. Thus once x is chosen, there are 8 choices for y. Thus |\mathsf{Aut}(Z_4 \times Z_4)| = 96.

Exhibit a presentation for the quaternion group

Prove that the following is a presentation for the quaternion group of order 8: Q_8 = \langle a, b \ |\ a^2 = b^2, a^{-1}ba = b^{-1} \rangle.


It suffices to show that some generating set of Q_8 satisfies these relations and that any group with this presentation has at most 8 elements.

Evidently, letting i = a and j = b, we have i^2 = -1 = j^2 and (-i)ji = (-i)(-k) = -j.

Now consider the presentation \langle a, b \ |\ a^2 = b^2, a^{-1}ba = b^{-1} \rangle. Note that a^2b^{-1} = 1, so that a^2(a^{-1}ba)(a^{-1}ba) = 1, so that ab^2a = 1. From this we deduce that a^4 = 1 and thus b^4 = 1.

  • There is one reduced word of length 0: 1.
  • There are two reduced words of length 1: a and b.
  • There are at most four reduced words of length 2: a^2, ab, ba, and b^2 = a^2. We see that at most three of these are reduced.
  • There are at most six reduced words of length 3: a^3, a^2b, aba = b, ab^2 = a^3, ba^2 = a^2b, and ba^2 = a^2b. We see that at most two of these are reduced.
  • There are at most four reduced words of length 4: a^4 = 1, a^3b = ba, a^2ba = ab, and a^2b^2 = 1. None of these are reduced.

Thus in fact Q_8 = \langle a, b \ |\ a^2 = b^2, a^{-1}ba = b^{-1} \rangle.

Exhibit a presentation for Sym(4) with two generators

Establish a finite presentation for S_4 with 2 generators.


Recall that S_4 is generated by \alpha = (1\ 2) and \beta = (1\ 2\ 3\ 4). Evidently, these elements satisfy the relations \alpha^2 = \beta^4 = (\alpha\beta)^3 = 1. We will now show that any group presented by \langle a, b \ |\ a^2 = b^4 = (ab)^3 \rangle has at most 24 elements.

  • There is one reduced word of length 0: 1.
  • There are two reduced words of length 1: a and b.
  • There are at most four reduced words of length 2: a^2 = 1, ab, ba, and b^2. We see that at most three of these are reduced.
  • There are at most six reduced words of length 3: aba, ab^2, ba^2 = b, bab, b^2a, b^3. We see that at most five of these are reduced.
  • There are at most ten reduced words of length 4: aba^2 = ab, abab, ab^2a, ab^3, baba = ab^3, bab^2, b^2a^2 = b^2, b^2ab, b^3a = abab, and b^4 = 1. At most five of these are reduced.
  • There are at most ten reduced words of length 5: ababa = b^3, abab^2, ab^2a^2 = ab^2, ab^2ab, ab^3a = bab, ab^4 = a, bab^2a, bab^3, b^2aba = bab^3, and b^2ab^2. We see that at most five of these are reduced.
  • There are at most ten reduced words of length 6: abab^2a, abab^3, ab^2aba = abab^3, ab^2ab^2, bab^2a^2 = bab^2, bab^2ab = ab^2a, bab^3a = b^2ab, bab^4 = ba, b^2ab^2a = ab^2ab^2, and b^2ab^3 = abab^2a. We see that at most three of these are reduced.
  • There are at most six reduced words of length 7: abab^2a^2 = abab^2, abab^2ab = b^2a, abab^3a = ab^2ab, abab^4 = aba, ab^2ab^2a = b^2ab^2, and ab^2ab^3 = ab^2ab. We see that none of these is reduced.

Thus any group with this presentation has at most 24 elements, and in fact we have S_4 = \langle a,b \ |\ a^2 = b^4 = (ab)^3 = 1 \rangle.

Exhibit a presentation of Alt(4) using two generators

Establish a finite presentation for A_4 using two generators.


We know that A_4 is generated by an element of order 2 and an element of order 3. Let \alpha = (1\ 2)(3\ 4) and \beta = (1\ 2\ 3). Evidently, these generators satisfy the relations \alpha^2 = \beta^3 = (\alpha\beta)^3 = 1. To show that these relations provide a presentation for A_4, it suffices to show that the group \langle \alpha, \beta \ |\ \alpha^2 = \beta^3 = (\alpha\beta)^3 \rangle has order at most 24. To that end, we will compute the reduced words in this group.

  • There is one word of length 0: 1.
  • There are two words of length 1: a and b. Both are reduced by definition.
  • There are at most four words of length 2: a^2, ab, ba, and b^2. Since a^2 = 1, it is not reduced. Thus 3 words of length 2 might be reduced.
  • There are at most six words of length 3: aba, ab^2, ba^2, bab, b^2a, and b^3. Since ba^2 = b and b^3 = 1, these words are not reduced. The remaining 4 words might be reduced.
  • There are at most eight words of length 4: aba^2, abab, ab^2a, ab^3, baba, bab^2, b^2a^2, and b^2ab. Since aba^2 = ab, abab = b^2a, ab^2a = bab, ab^3 = a, baba = ab^2, and b^2a^2 = b^2, these words are not reduced. The remaining two words might be reduced.
  • There are at most four words of length 5: bab^2a = b^2ab, bab^3 = ba, b^2aba = bab^2, and b^2ab^2 = aba. None of these are reduced.

Thus any group presented by \langle a,b \ |\ a^2 = b^3 = (ab)^3 = 1 \rangle has cardinality at most 12, and we have the presentation A_4 = \langle \alpha, \beta \ |\ \alpha^2 = \beta^3 = (\alpha\beta)^3 = 1 \rangle.

Classify the groups of order 1805

Show that the matrix A = \begin{bmatrix} 0 & -1 \\ 1 & 4 \end{bmatrix} is an element of order 5 in GL_2(\mathbb{F}_{19}). Use this matrix to construct a nonabelian group of order 1805 and give a prepentation for this group. Classify groups of order 1805. (There are 3 isomorphism types.)


We begin with a lemma.

Lemma: Suppose H = \langle S \rangle and K = \langle T \rangle are groups and \varphi : K \rightarrow \mathsf{Aut}(H) is a group homomorphism, and identify H and K as subgroups of H \rtimes_\varphi K. Then H \rtimes_\varphi K = \langle S \cup T \rangle. Proof: Note that if (h,k) \in H \rtimes K, then (h,k) = (h,1)(1,k). \square

We now move to the main results. Note first that 1805 = 5 \cdot 19^2.

It is straightforward to show that A^2 = \begin{bmatrix} -1 & -4 \\ 4 & -4 \end{bmatrix}, A^3 = \begin{bmatrix} -4 & 4 \\ -4 & -1 \end{bmatrix}, A^4 = \begin{bmatrix} 4 & 1 \\ -1 & 0 \end{bmatrix}, and A^5 = I.

We will now construct a nonabelian group of order 1805 as a semidirect product of Z_{19}^2 and Z_5 = \langle x \rangle. First, supposing Z_{19}^2 = \langle a \rangle \times \langle b \rangle, we write the element a^ib^j as the vector \begin{bmatrix} i \\ j \end{bmatrix}. Recall that \mathsf{Aut}(Z_{19}^2) \cong GL_2(\mathbb{F}_{19}); let \alpha \in \mathsf{Aut}(Z_{19}^2) be the image of the matrix A under the isomorphism \theta : GL_2(\mathbb{F}_{19}) \rightarrow \mathsf{Aut}(Z_{19}^2) given by \theta(X)(v) = Xv. By this previous exercise, there exists a unique group homomorphism \varphi Z_5 \rightarrow \mathsf{Aut}(Z_{19}^2) such that \varphi(x) = \alpha. Since \varphi is not trivial, Z_{19}^2 \rtimes_\varphi Z_5 is a nonabelian group of order 1805$.

Now we find a presentation for this group.

Now by the lemma, H \rtimes_\varphi K is generated by \mu = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \eta = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, and \omega = x. It remains to determine the pairwise products of these elements.

  1. Because H is abelian, we have \mu \eta = \eta \mu.
  2. Note that \omega \mu = (1,x)(\begin{bmatrix} 1 \\ 0 \end{bmatrix}, 1) = (A\begin{bmatrix} 1 \\ 0 \end{bmatrix}, x) = (\begin{bmatrix} 0 & -1 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}, x) = (\begin{bmatrix} 0 \\ 1 \end{bmatrix}, x) = \eta \omega.
  3. Note that \omega \eta = (1,x)(\begin{bmatrix} 0 \\ 1 \end{bmatrix}, 1) = (\begin{bmatrix} 0 & -1 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix}, x) = (\begin{bmatrix} -1 \\ 4 \end{bmatrix} x) = \mu^{-1} \eta^4 \omega
  4. Note that A \begin{bmatrix} -1 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}. Thus it is straightforward to show that \mu \omega = \omega \mu^{-1} \eta^4.

Thus this group has the presentation \langle \mu, \eta, \omega \ |\ \mu^{19} = \eta^{19} = \omega^5, \mu \eta = \eta \mu, \omega \mu = \eta \omega, \omega \eta = \mu^{-1} \eta^4 \omega, \mu \omega = \omega \mu^{-1} \eta^4 \rangle.

Now we classify groups of order 1805.

By FTFGAG, the abelian groups of order 1805 are (up to isomorphism) Z_{1805} and Z_{95} \times Z_{19}.

Now suppose G is a nonabelian group of order 1805. By Sylow’s Theorem, n_{19} = 1; let H \leq G be the unique (hence normal) Sylow 19-subgroup. Now let K = \langle x \rangle \cong Z_5 be a Sylow 5-subgroup of G; by Lagrange, H \cap K = 1, so that HK = G. By the recognition theorem for semidirect products, we have G = H \rtimes_\psi K for some \psi : K \rightarrow \mathsf{Aut}(H).

Now there are two groups of order 19^2 up to isomorphism: Z_{361} and Z_{19}^2. Suppose H \cong Z_{361}. Now \mathsf{Aut}(Z_{361}) \cong Z_{19 \cdot 18}, which has no element of order 5 by Lagrange. Thus every group homomorphism \psi : K \rightarrow \mathsf{Aut}(H) is trivial, and we have G \cong Z_{361} \times Z_5. But then G is abelian, a contradiction.

Thus we may assume that H \cong Z_{19}^2. Now \mathsf{Aut}(H) \cong GL_2(\mathbb{F}_{19}), and this group has order 123120 = 2^4 \cdot 3^4 \cdot 5 \cdot 19. In particular, the Sylow 5-subgroups of \mathsf{Aut}(H) have order 5. Now because \psi is nontrivial and Z_5 is simple, \mathsf{im}\ \psi is a Sylow 5-subgroup of \mathsf{Aut}(H), and thus (by Sylow’s Theorem) is conjugate to \mathsf{im}\ \varphi as constructed above. Since Z_5 is cyclic, by this previous exercise, Z_{19}^2 \rtimes_\psi Z_5 \cong Z_{19}^2 \rtimes_\varphi Z_5.

Thus the distinct groups of order 1805 are as follows.

  1. Z_{1805}
  2. Z_{95} \times Z_{19}
  3. Z_{19}^2 \rtimes_\varphi Z_5, where Z_5 = \langle x \rangle and \varphi(x) is any element of order 5 in \mathsf{Aut}(Z_{19}^2).

A criterion for the existence of unique group homomorphisms from an abelian group

Let A = \langle x_1 \rangle \times \cdots \langle x_t \rangle be a finite abelian group with |x_i| = n_i for each i.

  1. Find a presentation for A.
  2. Prove that if G is a group containing commuting elements g_1,\ldots,g_t such that g_i^{n_i} = 1 for each i, then there is a unique group homomorphism \theta : A \rightarrow G such that \theta(x_i) = g_i.

  1. We claim that A \cong B = \langle a_1, \ldots, a_t \ |\ a_i^{n_i} = 1, a_ia_j = a_ja_i \rangle, where 1 \leq i,j \leq t. Indeed, the “standard basis” elements e_i, consisting of x_i in the i-th coordinate and 1 in all other coordinates, satisfy these relations.
  2. Note that every element of A can be written uniquely as (x_i^{b_i})_{i=1}^t. Define \theta((x_i^{b_i})) = \prod_{i=1}^t g_i^{b_i}, where the product on the right hand side is as computed inside G. It is clear that \theta(x_i) = g_i for each i. Moreover, \theta is a homomorphism because the g_i commute pairwise.

    Finally, suppose \sigma is a homomorphism A \rightarrow G such that \sigma(x_i) = g_i for each i. Then \sigma((x_i^{b_i})) = \prod \sigma(x_i)^{b_i} = \prod \theta(x_i)^{b_i} = \theta((x_i^{b_i})). Thus \sigma = \theta (products computed inside G), so that \theta is unique.

Compute presentations for a given central product of groups

Give presentations for the groups Z_4 \ast_\varphi D_8 and Z_4 \ast_\psi Q_8 constructed in a previous example.


Note that Z_4 \times D_8 is generated by (x,1), (1,r), and (1,s). Hence Z_4 \ast_\varphi D_8 is generated by the images of these elements under the natural projection; moreover, the relations satisfied by these also hold in the quotient. We have an additional relation: note that (x,1)^2(1,r)^2 = (x^2,r^2) \in Z. Thus Z_4 \ast_\varphi D_8 = \langle a,b,c \ |\ a^4 = b^4 = c^2 = 1, a^2 = b^2, ab = ba, ac = ca, bc = cb^3 \rangle.

Similarly, Z_4 \ast_\psi Q_8 is generated by the natural images of (x,1), (1,i), and (1,j), and we have an additional relation because (x,1)^2(1,i)^2 = (x^2,-1) \in Z. Thus Z_4 \ast_\psi Q_8 = \langle a,b,c \ |\ a^4 = b^4 = c^4 = 1, a^2 = b^2 = c^2, ab = ba, ac = ca, bc = cb^3 \rangle

Find a group presentation for Cyc(n)

Find a presentation for Z_n with one generator.


We have Z_n = \langle x \ |\ x^n = 1 \rangle.