# Subgroup of a soluble group is soluble

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

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What is the definition of soluble group you're using? –  tomasz Dec 23 '12 at 21:06
@tomaz that it has a subnormal series where each factor is abelian. Is that not correct? –  hmmmm Dec 23 '12 at 21:48
It is. But another one which is also correct is that the derived series terminates in finitely many steps, as I have pointed out in my answer. :) –  tomasz Dec 23 '12 at 21:55
@tomasz Cool, thanks. I just looked forwards a couple of pages in my notes and that definition is there. I'll have a think about it that way as well, thanks :) –  hmmmm Dec 23 '12 at 22:08

I think everything you have is good up till the abelian part. For that I'd use the first isomorphism theorem to show that $(B\cap H)/(A \cap H)$ injects into $B/A$.

Let $\pi:B\rightarrow B/A$ be the canonical homomorphism and consider the restriction $\pi'=\pi\left|_{B\cap H}\right.$ Since $A\cap H\leqslant A$, we have that $A\cap H\leqslant \text{ker}(\pi')$, and conversely if $x\in \ker(\pi')$ then $x\in A$ so $x\in A\cap H$. Thus $(B\cap H)/(A \cap H)$ injects into $B/A$, so $(B\cap H)/(A \cap H)$ is abelian.

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Thanks, what did I do wrong in the last part? –  hmmmm Dec 23 '12 at 21:59
@hmmmm It's not wrong, just more complicated than necessary. –  Alexander Gruber Dec 23 '12 at 22:04
Ah yeah, I seem to be doing that quite a lot recently. Thanks for the answer –  hmmmm Dec 23 '12 at 22:09

Hint: a group is soluble iff its derived series terminates in finitely many steps.

If $H<G$, then what is the relationship between commutator subgroups of $H$ and $G$?

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@hmmmm: In fact, $H'\leq G'$. ;-) –  Babak S. Dec 25 '12 at 14:00