Question

Given a group G and let G′ be its commutator subgroup, that is, the minimal subgroup containing elements of the form xyx⁻¹y⁻¹

for x, y ∈ G. (Note that the set S := x, y ∈ G is not necessarily a subgroup).
Show that π(G′) = G′ for any automorphism π of G.

Answer 1

For any automorphism π of a group G, the image of the commutator subgroup under π is equal to the commutator subgroup. This can be proven by showing that π(G′) ⊆ G′ and G′ ⊆ π(G′).

Step-by-step explanation:

Let's prove that for any automorphism π of G, π(G′) = G′.

1. First, we need to show that π(G′) ⊆ G′. Let g′ be an element in G′. Then, g′ can be written as g′ = xyx⁻¹y⁻¹, where x, y ∈ G. Applying π to both sides, we have π(g′) = π(xy)π(x⁻¹y⁻¹) = π(x)π(y)π(x⁻¹)π(y⁻¹).
2. Since π is an automorphism, it preserves the group operation, so π(x)π(y)π(x⁻¹)π(y⁻¹) = π(x) π(y) ( π(x) )⁻¹ ( π(y) )⁻¹. Therefore, π(g′) is of the form xyx⁻¹y⁻¹, which means it is an element of G′.
3. Next, we need to show that G′ ⊆ π(G′). Let g′′ be an element in G′. Then, g′′ can be written as g′′ = xyx⁻¹y⁻¹, where x, y ∈ G. Applying π⁻¹ to both sides, we have π⁻¹(g′′) = π⁻¹(xy)π⁻¹(x⁻¹y⁻¹) = π⁻¹(x)π⁻¹(y)π⁻¹(x⁻¹)π⁻¹(y⁻¹).
4. Since π⁻¹ is also an automorphism, it preserves the group operation, so π⁻¹(x)π⁻¹(y)π⁻¹(x⁻¹)π⁻¹(y⁻¹) = π⁻¹(x) π⁻¹(y) ( π⁻¹(x) )⁻¹ ( π⁻¹(y) )⁻¹. Therefore, π⁻¹(g′) is of the form xyx⁻¹y⁻¹, which means it is an element of G′. Hence, π(G′) ⊇ G′.
5. Combining both parts, we can conclude that π(G′) = G′ for any automorphism π of G.