Question

Let G be a group containing subgroups M and N. Suppose that \(M\subseteq N_G(N)\)
Prove.
1) \(MN=\{mn|m\in M, n\in N\}\) is a subgroup of G
2) Prove that M, and N are subgroups of MN
3) Prove that N is a normal subgroup of MN
4) Let \(\varphi: M\rightarrow MN/N\) be the composition of group homomorphisms \(M\rightarrow MN\rightarrow MN/N\) Where the first one is inclusion map and the second one is canonical projection map. Prove that the kernel of \(\varphi\) is \(M\cap N\)
5)Prove that \(\varphi\) is surjective
6) Conclude that \(M/M\cap N \) isomorphic to \(MN/N\)
Please, help

Answer 1

1) starting from \(N_G(N)\) we have N is normal subgroup of \(N_G(N)\), M is subgroup of \(N_G(N)\). By second isomorphism theorem, we have MN is a subgroup of \(N_G(N)\), hence MN is a subgroup of G.

Answer 2

2) Let \(m \in M\), \(m = m*e_N \) then \(m \in MN\) that shows M is a subgroup of MN
similarly for \(n\in N\), we have M, N are subgroups of MN.

Answer 3

3) Let \(n \in N\), we need prove \((mn) n (mn)^{-1} \in N~~ \forall mn~~ \in MN\)
\((mn)n(mn)^{-1}= (mn)n(n^{-1}m^{-1})=mnm^{-1}\) \(\in N\) .
Since N is normal subgroup of \(N_G(N)\) hence for all element a in \(N_G(N)\) we have \(ana^{-1}\in N\)
Moreover \(M\subset N_G(N) \) (given), then if \(m\in M\) then, \(m\in N_G(N)\). Hence \(mnm^{-1}\in N\)
As shown, \(mnm^{-1}=(mn)n(mn)^{-1}\in N\) for all \(mn\in MN\)
Therefore, N is normal subgroup of MN.

Answer 4

4) Let \(\varphi: M\rightarrow MN/N\\~~~~~~~m \rightarrow mn N\)


\(i: M\rightarrow MN\\~~~~~m\rightarrow mn\) and \(\pi: MN\rightarrow MN/N\\~~~~~~~~mn\rightarrow mnN\)

then \(\varphi = \pi~ i\)

we need show kernel \((\varphi)= M\cap N\)


\(kernel (\varphi) = \{m \in M ~~| \varphi(m) = e_{MN/N}\}\)

Answer 5

\(\varphi (m)= \pi~ i(m) = \pi (im)= \pi (mn) = mn N\)
But \(\pi (mn) = mn N= (mN)(nN) = (mN) N \)
moreover, N is identity in MN/ N , hence \((mN)N= mN\)

Answer 6

But \(\varphi (m) = e_{MN/N}=N = mN\)
that shows \(m \in N\), but \(m\in M\) , hence \(m \in M\cap N\)
Hence \(kernel\varphi = M\cap N\)

Answer 7

@FibonacciChick666

Answer 8

connection issues :( ok give me a min to look

Answer 9

I stuck at part 5 and then 6

Answer 10

ok, can you explain the notation N_G(N), I just am not familiar with it

Answer 11

is it a cyclic?

Answer 12

It is a normalizer of N under G . It is defined by {g in G such that gNg^-1 = N}
Nope, G is just a group

Answer 13

ohh ok

Answer 14

ok, so for 5, my first step is to look at ways to express surjective. My gut is saying a proof by contradiction may work, let me bring in someone much better at this than me, but I can still try.

Answer 15

@ganeshie8 and @zarkon, Abstract algebra proofs

Answer 16

The function f is surjective if for every b∈Codom(f) there exists some a∈A such that f(a)=b
So, essentially, you will need to show that for any element of M, there is a corresponding value in MN/N

Answer 17

I have a theorem let me have \(\pi\) is surjective but we need \(\varphi = \pi i\)

Answer 18

so if we did something like M-->G and N--->G and N_g---->G

Answer 19

maybe?

Answer 20

Is it true that (surjective (injective)) = surjective ?