I'm trying to show that if $G$ is a soluble group with $H$ some subgroup then $H$ is also soluble. My argument is as follows:
As $G$ is soluble then we have the subnormal series:
$\{e\}\triangleleft G_1 \triangleleft..... \triangleleft G_n=G$.
If we now intersect $H$ with this series we get:
$\{e\}\triangleleft G_1\cap H \triangleleft G_2\cap H..... \triangleleft G_i\cap H\triangleleft H\cap G_i=H$
So we now need to show the normality and that each factor is abelian.
To see the normality we need to prove that given $A,B,H$ subgroups of $G$ such that $A\triangleleft B$ we have $A\cap H \triangleleft B\cap H$. So take $g\in A\cap H$ and $h\in B\cap H$ and consider $hgh^{-1}$.
Now as $h\in B$ and $g\in A$ we have $hgh^{-1}\in A$ also as $h\in H$ and $g\in H$ then $hgh^{-1}\in H$ and so we have that $hgh^{-1}\in A\cap H$ and this is normal.
Now we need to show that each factor is abelian. So we need to show that given $A,B,H$ subgroups of $G$ such that $A\triangleleft B$ we have:
If $B/A$ is abelian then $(B\cap H) / (A \cap H)$ is abelian. To see this we need to show that $(B\cap H) / (A \cap H)$ is a subgroup of $B/A$. So I am claiming that:
$(B\cap H) / (A \cap H)\cong A(B\cap H)/A$
Which is a subgroup of $B/ A$
Now we have the following, that $A\triangleleft B$ and $B\cap H < A$. So we apply the second isomorphism theorem to get:
$A(B\cap H)/A\cong (B\cap H)/ (A\cap H \cap B)=A(B\cap H)/A\cong (B\cap H)/ (A\cap H )$ as $A\cap B=A$
Is this correct, I am a bit worried about the last part.
Thanks very much any help