Groups

Answer key — Groups

1. 1.

   Disjoint cycles commute, so $\sigma^m = (1\,2\,3)^m (4\,5)^m$, and the two factors move disjoint sets of points, so $\sigma^m = e$ exactly when both factors are $e$, i.e. when $3 \mid m$ and $2 \mid m$. The least such $m$ is $\operatorname{lcm}(3, 2) = 6$. In general the order of a permutation is the lcm of the lengths of its disjoint cycles.

2. 2.

   The multiples of $8$ modulo $12$ are $8, 4, 0$: so $3 \cdot 8 = 24 \equiv 0$ and the order is $3$. In general $m k \equiv 0 \pmod n$ iff $n \mid mk$ iff $\frac{n}{\gcd(k,n)} \mid m$, so the order of $k$ is $n/\gcd(k, n) = 12/4 = 3$. The subgroup it generates is $\{0, 4, 8\}$, of order $3$ — which divides $12$, as Lagrange's theorem insists.

3. 3.

   By Lagrange's theorem the order of every subgroup divides $20$, whose divisors are $1, 2, 4, 5, 10, 20$. So $3$, $6$ and $8$ are impossible. Lagrange's theorem only rules orders out; it never guarantees that a divisor occurs — $A_4$ shows that a divisor ($6$ of $12$) may be missing. (For order $20$ every divisor does occur, but that needs Sylow's and Cauchy's theorems.)

4. 4.

   Lagrange's theorem says: if $H \le G$ then $|H|$ divides $|G|$. It does not say that every divisor is the order of a subgroup or of an element. Here the conclusion happens to be true — $(1\,2\,3)$ has order $3$ — and in general Cauchy's theorem guarantees an element of order $p$ for every prime $p$ dividing $|G|$. For composite divisors existence can fail: $A_4$ has order $12$ and no subgroup of order $6$.

5. 5.

   The normal subgroups are $\{e\}$, $A_3$ and $S_3$. $A_3$ is the kernel of the sign homomorphism $S_3 \to \{\pm 1\}$, hence normal (it also has index $2$). The subgroup $\{e, (1\,2)\}$ is not normal: conjugating by $(1\,3)$ gives $(1\,3)(1\,2)(1\,3) = (2\,3) \notin \{e, (1\,2)\}$. The same holds for the other two subgroups of order $2$.

6. 6.

   Let $G$ be the rotation group, acting on the faces. Any face can be rotated onto any other, so the orbit of a face $F$ has $12$ elements. A rotation fixing $F$ must turn it about the axis through its centre, by a multiple of $72^\circ$: the stabiliser has order $5$. By orbit–stabiliser, $|G| = 12 \cdot 5 = 60$. (Check with vertices: $20$ vertices, stabiliser of order $3$, $20 \cdot 3 = 60$; with edges: $30 \cdot 2 = 60$.) This group is isomorphic to $A_5$.

7. 7.

   By Burnside's lemma the number of orbits is $\frac{1}{24}\sum_g |\mathrm{Fix}(g)|$, and $|\mathrm{Fix}(g)| = 2^{c(g)}$ where $c(g)$ is the number of cycles of $g$ on the six faces. So

   $\frac{1}{24}\big(2^6 + 6 \cdot 2^3 + 3 \cdot 2^4 + 8 \cdot 2^2 + 6 \cdot 2^3\big) = \frac{64 + 48 + 48 + 32 + 48}{24} = \frac{240}{24} = 10.$

   Check by hand: with $0, 1, 2, 3, 4, 5, 6$ black faces there are $1, 1, 2, 2, 2, 1, 1$ colourings (two black faces are adjacent or opposite; three black faces meet at a corner or form a band), total $10$.

8. 8.

   Option (a). The theorem quantifies over all g : G and assumes H.index = 2. Option (b) is a triviality (closure under multiplication), (c) confuses order with index and is false, and (d) is the false converse. In the lesson, (a) is the key step in proving that $A_4$ has no subgroup of order $6$.

9. 9.

   from math import lcm
   
   def perm_order(p):
       seen = set()
       result = 1
       for start in range(len(p)):
           if start in seen:
               continue
           length, i = 0, start
           while i not in seen:
               seen.add(i)
               i = p[i]
               length += 1
           result = lcm(result, length)
       return result
   

   Why the lcm: disjoint cycles commute and move disjoint points, so $\sigma^m = e$ iff each cycle's $m$-th power is $e$, iff $m$ is a multiple of every cycle length. The least such $m$ is the lcm. The last test, cycle type $(2, 3, 5)$, gives $30$ — the largest order of an element of $S_{10}$.