Question

Let G be a group and let K and N be normal subgroups of G. Prove that there is an injective group homomorphism from
\(G/K\cap N\) to (G/K) x (G/N)
I have homomorphism part done but injective.
Please, help

Answer 1

@Alchemista This time, I carefully type the problem. :)

Answer 2

ya

Answer 3

someone tells me please, how to type x in latex such that it does not appear as \(x\)?

Answer 4

@zepdrix

Answer 5

cool

Answer 6

don't mess up my post , please,

Answer 7

I defined:
\[f: G/K\cap N \rightarrow (G/K)\times (G/N)\\~~~~g~(K\cap N)\rightarrow (gK, gN)\]

Answer 8

Let a, b \(\in G\), then \(a(K\cap N),~~b(K\cap N)\in K\cap N \)
and \(a(K\cap N) b(K\cap N)=ab(K\cap N)\)

Answer 9

\(\huge\tt \begin{align} \color{black}{x
}\end{align}\)

Answer 10

\(\huge\tt \begin{align} \color{black}{\times
}\end{align}\)

Answer 11

so that we have
\[f(ab (K\cap N) =(ab K, ab N)\]
but ab K = aK bK
and ab N = aN bN
so that it is \((aK, aN)(bK, bN)= f(a (K\cap N)*f(b(K\cap N)\)
that shows f is homomorphism

Answer 12

Show that the kernel is trivial.

Answer 13

show me, please

Answer 14

Show that the only element that gets sent to \((eK, eN)\) is the identity.

Answer 15

Is it not that we have to show f is one - to -one?

Answer 16

These are equivalent. Another way would be to show that
$$
f(a) =f(b) \implies a = b
$$

Answer 17

\(\begin{eqnarray*}
f(a) &=& f(b) \\
(aK, aN) &=& (bK, bN)
\end{eqnarray*}\)

Does this mean that \(a=b\)?

Answer 18

yes,

Answer 19

Make sure you clarify that this is true since they are equivalence class representatives.