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(3) Let G be an arbitrary group. For a, b ? G, the commutator [a, b] is the element: [a, b] = abab?¹ ? G. The commutator subgroup [G, G] is the subgroup of G generated by all commutators in G; so every finite product of any number of commutators is in [G, G]. For instance, if x, y, z, w ? G, then: [x,y][x,z][y, w] = (xyx?¹y?¹)(xzx?¹z?¹)(ywy?¹w?¹) = xyxyxzxzywyw?¹ ? [G,G]. (a) Prove that two elements a, b ? G commute if and only if [a, b] = e. (b) For a,b ? G, prove that the inverse of [a, b] can itself be written as a commutator. (c) For a,b,g ? G, prove that g[a, b]g?¹ can itself be written as a commutator. (d) Find the commutator subgroup [Q?, Q?] in the quaternion group Q?. (e) Prove that if G is abelian, then [G, G] = {e}. (f) Now (not assuming that G is abelian anymore), prove that [G,G] as defined above actually is a subgroup of G. (g) Prove that [G, G] is a normal subgroup of G. (h) Prove that the quotient group G/[G, G] is abelian. (This quotient group is often called the 'abelianization' of G)

          (3) Let G be an arbitrary group. For a, b ? G, the commutator [a, b] is the element:
[a, b] = abab?¹ ? G.
The commutator subgroup [G, G] is the subgroup of G generated by all commutators in G; so every
finite product of any number of commutators is in [G, G]. For instance, if x, y, z, w ? G, then:
[x,y][x,z][y, w] = (xyx?¹y?¹)(xzx?¹z?¹)(ywy?¹w?¹) = xyxyxzxzywyw?¹ ? [G,G].
(a) Prove that two elements a, b ? G commute if and only if [a, b] = e.
(b) For a,b ? G, prove that the inverse of [a, b] can itself be written as a commutator.
(c) For a,b,g ? G, prove that g[a, b]g?¹ can itself be written as a commutator.
(d) Find the commutator subgroup [Q?, Q?] in the quaternion group Q?.
(e) Prove that if G is abelian, then [G, G] = {e}.
(f) Now (not assuming that G is abelian anymore), prove that [G,G] as defined above actually
is a subgroup of G.
(g) Prove that [G, G] is a normal subgroup of G.
(h) Prove that the quotient group G/[G, G] is abelian.
(This quotient group is often called the 'abelianization' of G)

(3) Let G be an arbitrary group. For a, b ? G, the commutator [a, b] is the element:
[a, b] = abab?¹ ? G.
The commutator subgroup [G, G] is the subgroup of G generated by all commutators in G; so every
finite product of any number of commutators is in [G, G]. For instance, if x, y, z, w ? G, then:
[x,y][x,z][y, w] = (xyx?¹y?¹)(xzx?¹z?¹)(ywy?¹w?¹) = xyxyxzxzywyw?¹ ? [G,G].
(a) Prove that two elements a, b ? G commute if and only if [a, b] = e.
(b) For a,b ? G, prove that the inverse of [a, b] can itself be written as a commutator.
(c) For a,b,g ? G, prove that g[a, b]g?¹ can itself be written as a commutator.
(d) Find the commutator subgroup [Q?, Q?] in the quaternion group Q?.
(e) Prove that if G is abelian, then [G, G] = e.
(f) Now (not assuming that G is abelian anymore), prove that [G,G] as defined above actually
is a subgroup of G.
(g) Prove that [G, G] is a normal subgroup of G.
(h) Prove that the quotient group G/[G, G] is abelian.
(This quotient group is often called the 'abelianization' of G)

Added by Robert H.

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