*Algebra*

## Section 6.11. Permutation Representations

Exercise 6.11.1 An action of \(S_3\) on a set of four elements is a homomorphism \(\varphi : S_3 \to S_4\). The kernel of this homomorphism is a normal subgroup of \(S_3\). There are three cases:

- \(\ker\varphi = S_3\). The action \(\varphi\) is therefore trivial.
- \(\ker\varphi = A_3\). The image of \(\varphi\) has order 2, so it consists of the identity plus some element of \(S_4\) that has order 2, of which there are 9 in total (6 transpositions and 3 elements consisting of the product of a pair of disjoint transpositions). The even permutations in \(S_3\) therefore act trivially on the set, whereas the odd permutations all permute the set in the same way, acting as the other element of the image. There are therefore 9 such actions.
- \(\ker\varphi\) is trivial. Then \(\varphi(x)\) must have order 3 in \(S_4\); call it \(x'\). Likewise \(y' = \varphi(y)\) must have order 2, and we must have \(y'x' = x'^2 y'\) in order for \(\varphi\) to be a valid homomorphism. \(x'\) can only be a 3-cycle; without loss of generality, say \(x' = (1\ 2\ 3)\). We know there is one valid choice \(y' = (1\ 2)\) and if \(y'\) transposes two of the three elements 1, 2, 3 then we can always relabel 1, 2, 3 so that \(y' = (1\ 2)\). Now suppose \(y'\) is a transposition involving the index 4; without loss of generality, say \(y = (1\ 4)\) (we can simply relabel 1, 2, 3 to make this true). Then \(y'x' = (1\ 2\ 3\ 4)\) and \(x'^2y' = (1\ 4\ 3\ 2)\), so this choice of \(y'\) is not valid. Or suppose \(y'\) is a product of two transpositions; one of these must involve two of the three elements 1, 2, 3, so again we can simply relabel 1, 2, 3 so that \(y' = (1\ 2)(3\ 4)\). Then \(y'x' = (2\ 4\ 3)\) and \(x'^2y' = (2\ 3\ 4)\), so again this choice of \(y'\) is not valid. Therefore for any faithful action of \(S_3\) on a set of four elements there is some way of labelling the elements so that \(x\) acts as \((1\ 2\ 3)\) and \(y\) as \((1\ 2)\), \(S_3\) fixes one of the elements of the set and \(\sigma \in S_3\) acts on the other three elements by permuting them according to \(\sigma\).

Exercise 6.11.2 Suppose \(T\) acts on the set of two elements nontrivially. Then the action is a surjective homomorphism \(\varphi : T \to S_2\). But we argued in Exercise 6.9.4 that \(T\) is isomorphic to \(A_4\). We can now apply a similar argument to that used in Exercise 6.7.11: the image of a 3-cycle under \(\varphi\) must be the identity, but the 3-cycles generate \(A_4\), so \(\varphi\) is trivial, contradicting the assumption that \(\varphi\) is surjective. Therefore the only possible action of \(T\) on a set of 2 elements is trivial.

Exercise 6.11.3 Let \(\varphi : G \to \operatorname{Perm}(S)\) be the action of \(G\) on \(S\). Then, \(H\) is the subset of \(G\) consisting of elements mapped to the identity permutation; that is, \(H\) is the kernel of \(\varphi\). Therefore \(H\) is a normal subgroup of \(G\).

Exercise 6.11.4 \(D_4\) consists of the identity, three rotations, and four reflections. The three rotations each have no fixed points, each reflection across a diagonal has two fixed points, and the other two reflections, across lines parallel to two sides of the square, have no fixed point. That is, only the identity of \(D_4\) acts as the identity on the set of four vertices; this action is faithful. On the other hand, \(D_4\) does not act faithfully on the diagonals; for example, the 180 degree rotation fixes both diagonals.

Exercise 6.11.5 Denote the two orbits by \(\{a, b, c\}, \{d, e\}\). The image of the action \(\varphi : G \to \operatorname{Perm}(S)\) therefore consists only of those permutations that leave these two subsets invariant, that is, \(\im \varphi \subseteq S_3 \times S_2\). This restricts the possible actions to injective homomorphisms \(\varphi : G \to S_3 \times S_2\); in other words, \(G\) must be isomorphic to a subgroup of \(S_3 \times S_2\) (which will be its image under \(\varphi\)). We will identify \(G\) with this subgroup from now on, and represent its elements in the form \((g_1, g_2)\), where \(g_1 \in S_3\) and \(g_2 \in S_2\), and \(g_1\) acts on \(\{a, b, c\}\) while \(g_2\) acts on \(\{d, e\}\). Since \(G\) must act transitively on \(\{a, b, c\}\) and \(\{d, e\}\), the projections \(G_1 = \{g_1 \mid (g_1, g_2) \in G\}\) and \(G_2 = \{g_2 \mid (g_1, g_2) \in G\}\) must respectively act transitively on \(\{a, b, c\}\) and \(\{d, e\}\). This allows us to break down the possibilities into the following cases:

- \(G\) contains the element \((1, -1)\) where \(-1\) denotes the non-identity element of \(S_2\). This implies that \(G\) has the structure \(G_1 \times S_2\), where \(G_1\) is some subgroup of \(S_3\), and (as stated above) \(G_1\) acts transitively on \(\{a, b, c\}\). This implies that \(G_1\) is either \(S_3\) or \(C_3\), so \(G\) may be \(S_3 \times S_2\) or \(C_3 \times S_2\).
- \(G\) doesn't contain the element \((1, -1)\) and there is only one element \(\sigma \in S_3\) such that \((\sigma, -1) \in G\). Then there can't be any \(\tau \in S_3 \setminus \{1\}\) such that \((\tau, 1) \in G\), because the product would be \((\sigma', -1)\) where \(\sigma' \neq \sigma\). So \(G\) can only be \(\{(1, 1), (\sigma, -1)\}\), where \(\sigma\) has order 2. But this fails to satisfy the requirement that \(G_1\) acts transitively.
- \(G\) doesn't contain the element \((1, -1)\) and there are at least two elements, \(\sigma \neq \sigma'\), such that \((\sigma, -1) \in S_3\) and \((\sigma', -1) \in S_3\). Both \(\sigma\) and \(\sigma'\) must be transpositions; if \(\sigma\) were a 3-cycle, then \((\sigma, -1)^3\) would be \((1, -1)\), which contradicts our assumption. But any two distinct transpositions generate \(S_3\), so \(G\) must contain an element of the form \((g_1, g_2)\) for every possible \(g_1\), and \(g_2\) will be \(+1\) if \(g_1\) is even, \(-1\) otherwise. So \(G\) is just \(S_3\) together with an extraneous sign bit.

We see that the possibilities for \(G\) are \(S_3 \times S_2\), \(C_3 \times
S_2\) (*i.e.*, \(C_6\)), and \(S_3\), where the latter acts on
\(\{a, b, c\}\) in the usual way and according to its sign on \(\{d, e\}\). It
is easy to see that these groups yield the orbits \(\{a, b, c\}\) and
\(\{d, e\}\) as required.

Exercise 6.11.6 The subspaces are generated by (0, 1), (1, 0), (1, 1), and (1, 2). An invertible matrix will always map a one-dimensional subspace to a one-dimensional subspace and this mapping is injective. Therefore there exists a homomorphism \(\varphi : GL_2(F) \to S_4\) where each invertible matrix is mapped to the permutation it induces on these four subspaces. The kernel of this homomorphism consists only of diagonal matrices since only a diagonal matrix can both map (0, 1) to a multiple itself and also do likewise for (1, 0). Furthermore, such a diagonal matrix has to be \(\diag(1, 1)\) or \(\diag(2, 2)\) and not \(\diag(1, 2)\) or \(\diag(2, 1)\) as these would exchange the (1, 1) and (1, 2) subspaces. Since the kernel has order 2, the image of \(\varphi\) must have order \(|GL_2(F)|/2\). But \(|GL_2(F)| = (3^2 - 1)(3^2 - 3) = 48\), which can be determined by observing that an invertible matrix can have an arbitrary nonzero vector for its first column and then the second column can be anything except a multiple of the first column. So \(|\im \varphi| = 24\), which implies that the image is the entire group \(S_4\).

Exercise 6.11.7

- Suppose \(\varphi : D_4 \to S_n\) is injective. Then \(|S_n| \geq |D_4|\). This implies \(n \geq 4\). It was previously established in Exercise 6.11.3 that \(D_4\) acts faithfully on the vertices of a square, so \(n = 4\).
- Suppose \(\varphi : D_6 \to S_n\) is injective. Since \(D_6\) contains elements of order 6, \(S_n\) must as well. This implies that \(n\) is at least 5 (all elements of \(S_4\) have order 1, 2, 3, or 4). Now consider the action of \(D_6\) on the set \(\{E_1, E_2, E_3, F_1, F_2\}\) where the \(E\)'s are the three long diagonals of a hexagon, \(F_1\) is a set of three alternating sides of the hexagon, and \(F_2\) is the set of the other three sides. Any reflection in \(D_6\) acts as a transposition on \(E_1, E_2, E_3\), so no subsequent rotation, which acts as a 3-cycle on this set, can restore all three diagonals to their original positions. Among proper rotations, only the 180 degree rotation fixes the three diagonals, but it exchanges \(F_1\) and \(F_2\). Therefore the action we have defined is faithful, establishing that \(n = 5\).
- If \(\varphi : H \to S_n\) is injective, then \(\varphi(i)\) must have order 4, therefore it must contain a 4-cycle. Thus far we know \(n \geq 4\). Without loss of generality, say \(\varphi(i)\) contains the 4-cycle \((1\ 2\ 3\ 4)\). Then \(\varphi(-1)\) contains the transpositions \((1\ 3)\) and \((2\ 4)\). Now \(j^2 = -1\) also, so if \(j\) maps \(1\) to \(a\), then it must map \(a\) to 3, and if it maps 3 to \(b\), then it must map \(b\) to 1; that is, \(j\) must contain a 4-cycle \((1\ a\ 3\ b)\). It must also contain a 4-cycle \((2\ c\ 4\ d)\) by similar reasoning. If these were the same cycle, then the product of \(\varphi(i)\) and \(\varphi(j)\) would contain the transpositions \((1\ 3)\) and \((2\ 4)\). But this product is \(\varphi(k)\), and its square \(\varphi(-1)\) would then fix 1, 2, 3, and 4, which contradicts our earlier assumptions. Likewise if they were inverse cycles to each other then \(\varphi(k)\) would fix 1, 2, 3, and 4 and so would \(\varphi(-1)\). Therefore, we know that the two 4-cycles in \(j\) must be different, which requires \(n \geq 8\). The action of \(H\) on itself by left-multiplication is a faithful action on a set of size 8. Therefore \(n = 8\).

Exercise 6.11.8 We previously argued (see Exercise 2.4.6 and Exercise 2.6.10) that the automorphisms of \(C_n\) are precisely the functions \(\varphi_k\) defined by \(\varphi_k(x) = kx\) for \(k\) relatively prime to \(n\). In the case where \(n\) is a prime \(p\) there are therefore exactly \(p-1\) automorphisms, each sending 1 to a different nonzero value. \(\varphi_k\) are in bijective correspondence with the multiplicative group \(\mathbb{F}_p^\times\) which also contains exactly the elements \(1, 2, \ldots, p-1\). In fact, the set of automorphisms and the multiplicative group are isomorphic as groups, where the isomorphism sends \(k \in \mathbb{F}_p^\times\) to \(\varphi_k\). It is easy to verify that this is indeed a group isomorphism.