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 $Z_n = \langle x \ |\ x^n = 1 \rangle$.