where is erroneous?
The question is: List the cyclic Subgroups of U(30)
My answer {<1>,<7>,<13>,<17>,<19>,<23>,<29>} however the answer key tells me {<1>,<7>,<13>,<19>,<29>} (leaving out <17>,<23>)
Where am I erroneous?
Please help me.
The question is: List the cyclic Subgroups of U(30)
My answer {<1>,<7>,<13>,<17>,<19>,<23>,<29>} however the answer key tells me {<1>,<7>,<13>,<19>,<29>} (leaving out <17>,<23>)
Where am I erroneous?
ab=ba
a^(-1).ab.a^(-1) = a^(-1).ba.a^(-1)
(a^(-1).a) b^(-1) = a^(-1).b(a.a^(-1))
e.b^(-1) = a^(-1).b.e
b^(-1) = a^(-1).b
is that right?? I hope that I made it...
you mean that I have to make left and right..
make left:
ba^(-1) = a^(-1)b => aba^(-1) = aa^(-1)b
make right:
ba^(-1) = a^(-1)b => ba^(-1)a = a^(-1)ba
then what??
Show that whenever ab = ba, you have ba^(-1) = a^(-1)b.
then you said that
just answers:
For example, a left multiplication by LaTeX graphic is being generated. Reload this page in a moment.:
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Same idea: let d = ba^(-1). ?? I seem that not enough
ac=b
a (a^(-1)b)=b, if ac=ad=b, then c=d (left cancellation).. I m sure that's right.
da=b
d=ba(a^(-1)), if ad=ac=b, then d=c (right cancellation) is that right?
i m tried work on this..
GIVE a, b ∈ G
SHOW c, d ∈ G ?
ac=b da=b
ac=b ---> proof: a (a^(-1)b)=b (IS THAT RIGHT? I THINK SO AND THAT'S RIGHT)
da=b ---> proof: ?
I have tried set up matrices for that but not work. I don't know how to slove for this problem...
Let G be any group, and let a, b ∈ G. Show that there are c, d ∈ G such that ac = b and da = b. Hint: you have to give an explicit definition for c and d in terms of a and b.
Please I need your help for that qustion and how do slove that qustion's problem. can you help me for slove for that? Pleasee
Let G be any group, and let a, b ∈ G. Show that there are c, d ∈ G such that ac = b and da = b. Hint: you have to give an explicit definition for c and d in terms of a...