HINTS

Straightforward from linear algebra.

Mimick the proof for the left regular action.

1. Use the definition.
2. Show that 4 is in the orbit of 1 but 1 not in the orbit of 4.

Use that orthogonal maps preserve lengths. If the first column of a matrix has length 1, can it be an orthogonal matrix?

Try to find all matrices that fix the lines through the origin and (1,0), (0,1) and (1,1) respectively.

Straightforward computation.

Straightforward computation.

Use the following:

• (1,2,3,4)(1,2)(1,2,3,4)^-1 = (2,3)
• (1,2,3,4)(2,3)(1,2,3,4)^-1 = (3,4)
• (1,2,3,4)(3,4)(1,2,3,4)^-1 = (4,1)
• (1,2,3,4)(4,1)(1,2,3,4)^-1 = (1,2)

Straightforward.

• Do orbit calculations for the tetrahedron and the cube.
• There is one nontrivial permutation of the vertices of the cube that nevertheless fixes the diagonals (set wise). Why? Use this one to group the automorphisms of the cube in subsets of two that `induce' the same action on the diagonals.

Any automorphism of Q(a) has to fix 1 and thus Q. What equation should the image of a satisfy?

Try to rewrite the equations.

Try an inclusion map.

1. What happens if you conjugate a permutation (k, m)(r, s) with k, m, r, s distinct with an arbitrary element of S₄?
2. Compare multiplication tables and set up an isomorphism.
3. Choose 2 commuting elements of order 2, different from the ones in K.

If |H| = 3 then |Sym(G/H)| = 6, so the kernel of the morphism L has order at least 3.

1. Trivial!
2. Cycle structure.
3. Rewrite xy = yx.
4. Work out (xyx^-1)ⁿ.

Last question: If A commutes with every nonsingular n by n matrix then A commutes with every n by n matrix (check this). Let E_ij be the matrix with zeros everywhere except in the ij position. Now use that A commutes with E_ij.

Straightfoward.

S₄ has no elements of order 6. So what about S₄/K?

Straightforward computation.

Check the proofs of the classification of the groups of order at most 11.

The group contains an element of order 5. This element generates a normal subgroup. Prove this first.

Straightforward computation.

Straightforward computation.

A₂ is trivial.

Just do the checks.

Compare with the previous exercise.

Compare with the previous two exercises.

Do the checks.

Last question: Orbits are now the sets of vectors of fixed (even) weight.

Straightforward.

For every two bases of R³ we can find a linear transformation sending one to the other.

Do this computation.

What is the definition of orbit?

Draw a picture.

Since G is transitive there is a g G with g(x) = x'. This g works.

Just do some clever computations.

Straightforward.

Straightforward checking.

Use that a normal subgroup is a union of conjugay classes.

Straightforward.

Straightforward computations.

To prove that H is normal in G it suffice to prove that for all g in G and h in H the element g^-1hg is in H. Fix a g in G and an h in H. Since G is a finite group, every element from G, in particular g, is a finite product of elements from B. So, g can be written as

with the b_i in B.

Similarly, every element from H is a finite product of elements from H and h can be written as

with the a_i in B.

Prove by induction on n that g^-1hg is in H.

Just use the definition.

• The two cosets H and gH partition G.
• What is the kernel of the representation?

G contains an element of order p. It generates a subsgroup H with index 2. Now follow the steps.

Notice that for g in G and x in X we have gx^-1g^-1 = (gxg^-1)^-1. Hence without loss of generality we may assume that the set X is closed under taking inverses. Now use that every element y in <X> is a product of elements from X.

Use that normal subgroups are unions of conjugacy classes.

Straightforward checking of all the properties of a subgroup. For item 5: Formulate in terms of the centraliser when two conjugates of g are the same.

In each case, give a morphism such that the image is H and the kernel is N.

1. Each s-tuple is part of a t-tuple.
2. If n - 2 elements of a permutation are already fixed, can you turn the permutation into an even one?

1. It suffices to show that G contains all transpositions.
2. It suffices to show that G contains all 3-cycles.

Straightforward.

Show that every move is in A₆. Then do an order computation: First compute the orbit of 1, then consider the stabilizer of 2 and the orbit of 2, etc.

Show that this group is transitive on the vertices, then show that the stabilizer of 1 is transitive on the remaining vertices, etc. In particular, show that G is transitive on ordered 4-tuples. Use this to show that G contains all 4-cycles. Then show that G contains a 3-cycle by taking the product of two suitable 4-cycles. Conclude that G contains all 3-cycles and an odd element.