# Mathematics 602 Homework (due Apr 21) 15) Hint: ```Mathematics 602
Homework (due Apr 21)
A. Hulpke
15) Show that G = ⟨x, y ∣ x 2 , y2 ⟩ is infinite.
Hint: Find a group (e.g. in GL(2, Q)) that must be a quotient of G, but contains elements of infinite order.
16) A word w = a1 ⋯a n is called cyclically reduced if a1 =/ a −1
n . For a given word w, we denote by σ(w) the
cyclically reduced word obtained by deleting a prefix and postfix if the word is not cyclically reduced. Show
that two elements u, v of a free group F are conjugate in F if and only if there are words w1 , w2 such that
σ(u) = w1 w2 and σ(v) = w2 w1 .
17) a) Determine a presentation for S3 on the generators a = (1, 2, 3) and b = (2, 3).
b) Determine a presentation for S3 on the generators a = (1, 2, 3), b = (2, 3) and c = (1, 2). (Hint: Use relators
to express the new generator in the old ones.)
c) Determine a presentation for S3 on the generators c = (1, 2) and b = (2, 3).
d) Using that S4 = C22 ⋊ S3 , determine a presentation for S4 .
18) Show that every finitely generated group G has only finitely many subgroups of index n for any given n.
Hint: Every subgroup of index n gives rise to a homomorphism G → S n , described completely by the
generator images. Show that there are just finitely many possibilities for generator images.
Can you find a counterexample of an infinitely generated G?
19) (J. P. Serre, [MSWZ93]) Let F be the free group on 26 generators a, b,. . . . We create relations l = r if
l and r are English words that sound the same but are spelled differently (assume “news reader” English)
(for example see = sea). What can you tell about the structure of the finitely presented group given by these
relations?
20) Let G = ⟨a, b, c ∣ b a = b2 , c b = c 2 , a c = a 2 ⟩ and S = ⟨a, b a ⟩. Perform by hand a coset enumeration for S
as a subgroup of G. Use the result to conclude that G must be trivial.
21) Show that for every prime p &gt; 2 there are exactly two nonisomorphic nonabelian groups of order
2
p3 , namely the group ⟨x, y ∣ x p = y p = 1, x y = x p+1 ⟩ (which has a cyclic normal subgroup of order p2 ) and
⟨x, y, z ∣ x p = y p = z p = 1, y x = yz, z x = z y = 1⟩ (in which all elements have order 1 or p, but which is not C 3p ).
Hint: You know from M566 (G/Z(G) cyclic implies G abelian) that ∣Z(G)∣ = p and G/Z(G) ≅ C 2p .
Computer Problems The following problems are best done with use of a computer, e.g. GAP. Section III.12
in the “Notes on Computational Group Theory” describes basic commands. For example:
gap&gt; f:=FreeGroup(&quot;a&quot;,&quot;b&quot;);;
gap&gt; rels:=ParseRelators(f,&quot;a2,b3,(ab)5&quot;);
[ a^2, b^3, (a*b)^5 ]
gap&gt; g:=f/rels;
&lt;fp group on the generators [ a, b ]&gt;
gap&gt; AssignGeneratorVariables(g);
#I Global variable ‘a’ is already defined and will be overwritten
#I Assigned the global variables [ a, b ]
gap&gt; u:=Subgroup(g,[a,a^b]);
Group([ a, b^-1*a*b ])
gap&gt; Index(g,u);
6
gap&gt; CosetTable(g,u); # transposed, rows correspond to a,a^-1,b,b^-1
[ [ 1, 4, 3, 2, 6, 5 ], [ 1, 4, 3, 2, 6, 5 ], [ 2, 3, 1, 5, 6, 4 ], [ 3, 1, 2, 6, 4, 5 ] ]
gap&gt; AbelianInvariants(g); # result [2,3,3] would indicate Z_2 x Z_3 x Z_3
[]
gap&gt; LowIndexSubgroupsFpGroup(g,5);
[ Group(&lt;fp, no generators known&gt;), Group(&lt;fp, no generators known&gt;) ]
gap&gt; List(last,x-&gt;Index(g,x));
[ 1, 5 ]
gap&gt; Size(g);
60
22) Select generators for S5 and find sufficiently many identities amongst the generators to create a presentation for S5 . You may use GAP to verify that the group given by your presentation has order 120.
23) Let G = ⟨a, b, c ∣ a2 = b3 = c 4 = (ab)2 = (ac)2 = (bc)3 = [c 2 , b] = 1⟩. (Note: [x, y] = x −1 y −1 x y is called
the commutator of x and y.)
Let S = ⟨a, c⟩ ≤ G. By enumerating the cosets of S in G, determine permutations for the action of G on the
cosets of S.
24) Let H = ⟨a, b ∣ a 2 = b3 = (ab)7 ⟩ (this group is called the Hurwitz Group). Using a low index algorithm,
show that:
a) H has no proper subgroups if index &lt; 7.
b) PSL2 (7) can be generated by two elements x, y with ∣x∣ = 2, ∣y∣ = 3, ∣x y∣ = 7.
c) n = 15 is the smallest value of n such that A n can be generated by elements x, y with ∣x∣ = 2, ∣y∣ = 3, ∣x y∣ = 7.
25) Let G = ⟨a, b ∣ aba −2 bab −1 , (b−1 a3 b−1 a−3 )2 a⟩. Using the Low-Index algorithm as implemented in
GAP, find a subgroup S ≤ G such that [S ∶ S ′ ] is infinite (test AbelianInvariants for S), and thus G
is infinite.
References
[MSWZ93] Jean-François Mestre, René Schoof, Lawrence Washington, and Don Zagier.
homophones des groupes libres. Experiment. Math., 2(3):153–155, 1993.
Quotients
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