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11:01 AM
@robjohn hi
 
 
1 hour later…
12:24 PM
@JackRod: Did you have a question?
 
12:52 PM
@robjohn yes
sir how we do premutation fo groups
like I have a question where $\alpha=(1,6,7,4)(2,3),(8,5)$
and $\beta=*(1,7,3,)(2,6)(4,5,8)$
I need to find $\alpha^{-1} \beta\alpha$
 
1:32 PM
$(1,3)(2,4,6)(5,7,8)$
to see where $1$ goes, you apply $\alpha$, getting $6$, then apply $\beta$, getting $2$, then apply $\alpha^{-1}$, getting $3$
then $3\overset\alpha\to2\overset\beta\to6\overset{\alpha^{-1}}\to1$, so you get the cycle $(1,3)$
$2\overset\alpha\to3\overset\beta\to1\overset{\alpha^{-1}}\to4$
$4\overset\alpha\to1\overset\beta\to7\overset{\alpha^{-1}}\to6$
$6\overset\alpha\to7\overset\beta\to3\overset{\alpha^{-1}}\to2$
getting the cycle $(2,4,6)$
 
 
4 hours later…
5:13 PM
@robjohn hi
sir do u apply this
I meant how u apply them
can u teach me
 
5:52 PM
Do you understand what $(1,3,4)$ does?
$1\to3$, $3\to4$, and $4\to1$.
 
6:05 PM
hi
permutation of a group
@robjohn
 
You have to index the elements of the set
then $g_k$ gets replaced by $g_{\sigma(k)}$
 
yes sir cyclic notation
 
so $\sigma(g_k)=g_{\sigma(k)}$
 
ok got it
ok got it
 
so you only have to really work with the indices (integers)
 
6:16 PM
sir there is one more question?
1 Consider U(15) = {1, 2, 4, 7, 8, 11, 13, 14} under multiplication modulo 15.
This group has order 8. To find the order of the element 7, say, we compute the sequence 7^1
first of all what is U(15)
 
is it not the set you have written?
 
yes
but from where 15 comes
 
I usually see it written as $\mathbb{Z}_{15}^\ast$
the multiplicative group mod $15$
In modular arithmetic, the integers coprime (relatively prime) to n from the set { 0 , 1 , … , n − 1 } {\displaystyle \{0,1,\dots ,n-1\}} of n non-negative integers form a group under multiplication modulo n, called the multiplicative group of integers modulo n. Equivalently, the elements of this group can be thought of as the congruence classes, also known as residues modulo n, that are coprime to n. Hence another name is the group of primitive residue classes modulo n. In...
it consists of the integers that are relatively prime to $n$
 
ok got it
for order of element 7
book we need to compute g^1,g^2,till it gives 1
that will be the order of element
is it multiply of g
I mean 7?
 
yes, the order of an element, $g$, is the smallest positive integer, $k$, so that $g^k=1$
 
6:28 PM
but sir how can the 7^1=4
 
what?
 
book says order of the element 7 is 4
that is 7^4=1
 
@JackRod there, that looks better
$7^2=49\equiv4\pmod{15}$
$7^4\equiv4^2=16\equiv1\pmod{15}$
so $7^4\equiv1\pmod{15}$
 
oh
 
you could, of course, just compute $7^4=2401$ and then $2401\equiv1\pmod{15}$.
 
6:42 PM
yes sir got it now
 
o/
 

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