# Dummit & Foote 3.2.18 (and the second isomorphism theorem?)

I am self-studying my way through Dummit and Foote's Abstract Algebra (3rd Ed), and am currently working my way through the following problem (Exercise 18 in Section 3.2):

Let $$G$$ be a finite group, let $$H$$ be a subgroup of $$G$$, and let $$N\trianglelefteq G$$. Prove that if $$|H|$$ and $$|G:N|$$ are relatively prime then $$H\le N$$.

After wrestling with this for a little while, I feel like I might have made a breakthrough that would help my broader understanding greatly, but I would appreciate some feedback on whether I am being blinded by my small sample size of examples.

Consider $$G=S_4$$ and $$N=\langle(12)(34),(13)(24)\rangle$$. Since $$N$$ is normal, we can make a nice table of the left cosets and know that each column has a well-behaved product of its own, like this.

Now, if I highlight the members of any $$H\le G$$, I find that every column has either nothing highlighted or exactly $$|H\cap N|$$ cells highlighted. For instance,

• $$H=A_4$$: the leftmost column and rightmost two columns are highlighted
• $$H=\langle(12),(12)(34)\rangle$$: the leftmost two columns are highlighted
• $$H=\langle(12),(13)\rangle$$: one cell from each column is highlighted (they do not form a rectangle in my diagram, but that could be rectified if I chose my left coset representatives with more care)
• $$H=\langle(13),(24)\rangle$$: there are two columns with two highlighted cells apiece (see below)

Based on the way things these pictures are working in $$S_4$$, it looks like every subgroup can be arranged as a highlighted rectangle whose height and width are factors of the height and width of the entire grid. Proving the conjecture seems to be easier than finding it if I'm translating it correctly, because the statement about rows is saying $$|H\cap N|$$ is a factor of $$|N|$$ and the statement about columns is saying $$|H:H\cap K|=|HN:N|$$ is a factor of $$|G:N|$$ and those are both pretty obvious since $$H\cap N\le N$$ and $$HN\le G$$.

With that lemma out of the way, the problem at the top is easy to solve. In the context of my diagram, if $$|H|$$ (the area of the highlighted area) and $$|G:N|$$ (the number of columns in the entire grid) are relatively prime, then the highlighted $$H$$ can only lie in the single leftmost column, so $$H\subset N$$ and thus $$H\le N$$. That all comes out of Lagrange's theorem, then.

Is this a valid argument? Generally, D&F are much better with the hand-holding in the exercises, but I really needed to go on a journey here. But if I'm right, it's giving me a handle on the Second Isomorphism theorem that I didn't have in my first few times trying to tackle it.

Your argument seems to be trying to work with a coset/Cayley table. If I understood you correctly, you seem to be reaching the right conclusions, but it is difficult for me to understand your reasoning so I cannot say if that reasoning is on solid ground.

Is $$|H\cap N|$$ a factor of $$|N|$$?

Yes; but this does not depend on $$N$$ being normal, just on $$H$$ and $$N$$ being subgroups (and $$N$$ being finite).

Why does this hold? Well remember that:

Lemma. If $$G$$ is a group, and $$A$$ and $$B$$ are both subgroups of $$G$$, then $$A\cap B$$ is also a subgroup of $$G$$. In particular, since it is contained in both $$A$$ and $$B$$, it is also a subgroup of both $$A$$ and of $$B$$.

So: $$H\cap N$$ is a subgroup of $$N$$. But by Lagrange's Theorem (which you mention, so you must know it), the order of a subgroup of a finite group always divides the order of the group. So $$|H\cap N|$$ divides $$|N|$$, which is what you were asking about.

Is $$|HN:N|$$ a factor of $$|G:N|$$?

Yes. Here we can use that $$N$$ is normal (you don't strictly need it, but you can use it): note that $$HN/N$$ is a subgroup of $$G/N$$, and so again by Lagrange's Theorem we know that $$|HN/N| = |HN:N|$$ divides $$|G/N|=|G:N|$$.

$$|H:H\cap N|=|HN:N|$$?

I'm not clear if you are actually asking this or you know it; this is true for any two subgroups when the quantities are finite, since for subgroups $$A$$ and $$B$$, we always have $$|A|\,|B|=|AB||A\cap B|.$$ Thus, $$|H||N|=|HN||H\cap N|$$, so $$|H|/|H\cap N| = |HN|/|N|$$, and the result about indices follows (since these are all quantities, so $$|H:H\cap N|=|H|/|H\cap N|$$, etc.

You say "With that lemma out of the way..." I'm not sure what "Lemma", since you never enunciate one.

But these results now give you what you want: Note that $$|H:H\cap N|=|HN:N|$$ divides $$|H|$$, because $$|H:H\cap N| = |H|/|H\cap N|$$; on the other hand, it also divides $$|G:N|$$. Thus, $$|H:H\cap N|$$ divides $$\gcd(|H|,|G:N|)$$.

But we are assuming that $$\gcd(|H|,|G:N|)=1$$. That means that $$|H:H\cap N|=1$$. The only way this can occur is if $$H=H\cap N$$, which in turns means that $$H\subseteq N$$, as desired.