- #1
cmj1988
- 23
- 0
Given a group G. J = {[tex]\phi[/tex]: G -> G: [tex]\phi[/tex] is an isormophism}. Prove J is a group (not a subgroup!).
Well we know the operation is function composition. To demonstrate J a group we need to satisfy four properties:
(i) Identity: (I'm not sure what to do with this)
(ii) Inverses: Suppose a [tex]\phi[/tex] in J. We know that since [tex]\phi[/tex] is a one-to-one and onto function, we know that it has an inverse (demonstrated in class). Hence we know for every [tex]\phi[/tex] in J, there exists a [tex]\phi[/tex]^-1.
(iii) Closure: Suppose a [tex]\phi1[\tex] and [tex]\phi2[/tex] such that [tex]\phi[/tex]1(g)=g and [tex]\phi2[/tex](g)=g. Therefore [tex]\phi[/tex]1([tex]\phi2[/tex](g))=g. This follows from what we proved about functions in class.
(iv) Associativity: This one is just moving a bunch of parentheses aroudn while composing and hoping (fingers crossed) that regardless of order of composition a g pops out.
My big problem is identity.
Well we know the operation is function composition. To demonstrate J a group we need to satisfy four properties:
(i) Identity: (I'm not sure what to do with this)
(ii) Inverses: Suppose a [tex]\phi[/tex] in J. We know that since [tex]\phi[/tex] is a one-to-one and onto function, we know that it has an inverse (demonstrated in class). Hence we know for every [tex]\phi[/tex] in J, there exists a [tex]\phi[/tex]^-1.
(iii) Closure: Suppose a [tex]\phi1[\tex] and [tex]\phi2[/tex] such that [tex]\phi[/tex]1(g)=g and [tex]\phi2[/tex](g)=g. Therefore [tex]\phi[/tex]1([tex]\phi2[/tex](g))=g. This follows from what we proved about functions in class.
(iv) Associativity: This one is just moving a bunch of parentheses aroudn while composing and hoping (fingers crossed) that regardless of order of composition a g pops out.
My big problem is identity.