Recent content by nobody56

Yea, but I wasnt sure if I could apply the same rules to a map. In my book it just has for order is of a group or the order on an element. Thanks for your help and for clarifing that.

are the other two parts ok? and you said 5 because... \phi((a,b))=(-b,a) \phi((-b,a))=(-a,-b) \phi((-a,-b))=(b,-a) \phi((b,-a))=(a,b) \phi((a,b))=(-b,a) right? (I can not find anything to do with order of maps in my book)

Homework Statement Let G={(a,b)/ a,b\inZ} be a group with addition defined by (a,b)+(c,d)=(a+c,b+d). a) Show that the map\phi:G\rightarrowG defined by \phi((a,b))=(-b,a) is an automorphism of G. b) Determine the order of \phi. c) determine all (a,b)\inG with \phi((a,b))=(b,a)...

i decided to play it safe, and just wrote it all out, thank you for your time and help!

So then would i just be able to say that since CD4(r2)={1, r2, s, sr2},and CD4(s)={1, r2, s, sr2}, and CD4(sr2)={1, r2, s, sr2}, then CD4(1, s, r2, sr2)={1, r2, s, sr2}...or should i try and show each part, as in 1*r=r*1, which seems kinda tidious

right, so for b could I take the definition of a Centralizer CG(H)={g element of G, such that ga=ag} and just plug in the elements to show CG(H)=H? and do the same for the normalizer?

yeah, i forgot the ordering...as for c^(-1)Hc=(c^(-1)Hc)^(-1)=cHc^(-1) by the same ordering property right?

could i say let a, b be elements of NG(H), and let a = aHa^-1 and b = bHb^(-1), and since a and b are elements in NG(H) aHa^(-1)=bHb^(-1), then use the division algorithm for right cancellation to say aHa^(-1)b=bHb^(-1)b which goes to aHc^(-1)=bHe and then similarly by the division algorithm for...

Would c^(-1)=(ab^(-1))^(-1)=a^(-1)b?

[b]1. Let G be a Group, and let H be a subgroup of G. Define the normalizer of H in G to be the set NG(H)= the set of g in G such that gHg-1=H. a) Prove Ng(H) is a subgroup of G b) In each of the part (i) to (ii) show that the specified group G and subgroup H of G, CG(H)=H, and NG(H)=G...