Question

Let N be a subgroup of a group G. Prove TFAE
a) N is normal in G
b) for each a in G, there exists b in G such that aN=Nb
c) each right coset of N is also a left coset
d)For each a in G, aN = Na
Please, help

Answer 1

Hi Loser

Answer 2

To me, it is trivial and it is definition of a normal subgroup, so that it is hard to prove
a) --> b) is ok, there is something to say,
b)--> c) ?
c)--> d) is trivial, is it not that if right coset = left coset, then aN = Na? why such that question is made?

Answer 3

Hi!! Preetha

Answer 4

What is TFAE?

Answer 5

the following are equivalent

Answer 6

Ah! Thanks.

Answer 7

Just pretending to help. (-:

Answer 8

d) --> a) is the definition of normal subgroup. How to prove the definition? ha!!

Answer 9

will solve it to u in morning ;)
now i have to sleep , its more theoretical thing not much to prove :O

Answer 10

ok lets start from a :-
assume a is true
Let N be a subgroup of a group G and N is normal in G ,according to definition of normal subgroups aH=Ha for all a in G .
* since \(a \in G\) then let \(a=b \) thus for all a in G there exist b s.t aN=Nb
Hence b) holds .
** let aN be a right coset , according to a) aH=Ha take a^-1 for both sides
aNa^-1=Naa^-1
aNa^-1=Ne
aNa^-1=N .. thus each right coset of N is also a left coset
*** from definition

Answer 11

now let b is true ,Let N be a subgroup of a group G , for each a in G, there exists b in G such that aN=Nb.
* aN=Nb take b^-1
aNb^-1=N take a inverse
Nb^-1=a^-1 N
since aN=Nb and a^-1=Nb^-1 then a=b
thus aN=Na for a in G , thus N is normal
**aN=Nb , aH is right coset , Hb is left coset thus aN is also left coset and Nb is also right coset thus each right coset of N is also a left coset .
***from * u can also conclude this :)

Answer 12

now let c be true ,each right coset of N is also a left coset
which means for all aN in G ,aN=Nb ( b holds *)
continue the same way as for part b :)

Answer 13

now let d be true ,For each a in G, aN = Na then C hold then B hold then a hold xD

Answer 14

hope u got it , if u have any doubt just ask

Answer 15

@ikram002p from b to c
I don't get
\[aN= Nb\\aNb^{-1}=Nbb^{-1}=N\\a^{-1}aNb^{-1}=aN\\Nb^{-1}=aN\\aN=Nb=Nb^{-1}\]
it didn't lead to a =b

Answer 16

u copied it wrong :O
see if now make sense :-
aN=Nb
a^-1aN=a^-1Nb
N=a^-1 Nb

aNb^-1=N

a^-1N b = a N b^-1 from N=N

to make this true a need to be b
its something u can conclude from cosets properties not algebraically .

Answer 17

"to make this true , a need to be b"
that is \(a^{-1} N a = aNa^{-1}\)
or else?

Answer 18

yes only this :)

Answer 19

but we don't have this condition, right?

Answer 20

remember
aN=Na iff N=a^-1Na
this is the condition :)

Answer 21

it is aNa^-1, not a^-1 Na

Answer 22

both are equivalents :O but yep u have a point

Answer 23

ok, Thanks for solution, I 'll check other condition. If I don't get something, I will nag you again, :)

Answer 24

hehe anytime :)
i'll also see if i got something better , since u seems have doubts
and i'll post if i could :D

Answer 25

Honestly, I am not convinced by this logic. Please, check this logic. I use properties of cosets
aN = Nb
left cosets has property: aN = bN ,
then aN = bN = Nb
bN = Nb show that leftcoset = right coset, since a , b is arbitrary, then aN = Na

Answer 26

@dan815 look at the last attachment.

Answer 27

-.-