Showing There is No Element of Some Order in a Group
26 Aug 2021
I took a practice algebra qual the other day, and was totally stumped by a pretty basic group theory question:
Assume that $G$ is a simple group of order $168$.
(a) How many elements of order $7$ are there in $G$?
(b) Show that $G$ does not contain any elements of order $14$.
Part (a) is a pretty routine application of the Sylow theorems^{1}, but part (b) stumped me. I don’t think I’ve ever seen a problem like tihs before, and it’s (at least to me) entirely unobvious what to do. Of course, we discussed it after the mock qual, and after meditating on the solution for a bit, I’m writing this up so I’ll hopefully remember the idea.
$\ulcorner$
Let $H \leq G$ with $H = 14$. Then $H$ has a $7$sylow subgroup $P$, and since $[H:P] = 2$, we must have $P \vartriangleleft H.$ So $H$ is contained in $N_G(P)$ (the normalizer^{2} of $P$).
Now we use a fact that I had entirely forgotten: \([G : N_G(P)] = n_p = \# p\text{sylow subgroups}\). Once you see this, it becomes obvious (exercise: why?^{3}), and I’m going to try to not forget it again.
With this in hand, though, the rest comes fairly quickly. If $[G : N_G(P)] = n_7 = 8$, then
\[N_G(P) = \frac{G}{[G : N_G(P)]} = \frac{168}{8} = 21.\]But we know $H \leq N_G(P)$, so \(H \, \bigg \vert \, N_G(P)\) and we find $14 \, \big \vert \, 21$, giving us the desired contradiction.
$\lrcorner$
The most obvious way to show that $H \not \leq G$ is to use lagrange’s theorem, and this is still using lagrange’s theorem, just in a slightly sharper way.
Normally, we show that $H$ does not divide $G$, but if we can control $H$ enough to guarantee $H \leq K \leq G$, then we can actually apply lagrange’s theorem to this subgroup instead! As we saw above, $G$ was not restrictive enough to prevent a subgroup of order $14$, but the smaller group $N_G(P)$ was able to get the job done!
Now, how could we have been clued into thinking about $H \leq N_G(P)$? Well, primed by the previous paragraph we notice that $14$ does divide the order of $G$. So if we’re going to show the nonexistence of this subgroup, we’ll need something stronger. Without knowing how $H$ embeds into $G$, though, we have very few tools for finding a $K$ with $H \leq K \leq G$. In fact, we really only have two tools which let us move up in the subgroup lattice.
 Taking normalizers (note $H \leq N_G(H)$ for every $H$)
 Working with $p$subgroups
The normalizer angle isn’t promising, since we don’t know how $H$ should sit inside $G$, so we don’t know anything about how it conjugates. The $p$subgroup idea seems reasonable, though. After all, every $p$subgroup is contained in a Sylow $p$subgroup. So we might be able to pass information from the bottom of the subgroup lattice to the top.
This is doubly effective in our current case, since $7$ is the maximal power of $7$ in both $H$ and $G$. So actually the (unique!) $7$sylow subgroup of $H$ is already one of the $7$sylow subgroups of $G$. A very common idea with $p$groups is to take normalizers^{4}, indeed if you’ve done much finite group theory, you’re surely familiar with the frattini argument. The standard example of this argument is showing
If $H \vartriangleleft G$, and $P$ is a $p$sylow subgroup of $H$, then $G = N_G(P) H$.
So when we see $p$sylow subgroups, we might be trained to think about their normalizers. But as soon as we look at the normalizer of $P$, we see it contains $H$, and we can finish the proof from there.
As a quick epilogue, I asked mse for general techniques for showing $G$ does not have a subgroup of order $k$. There have been some comments, but nothing substantial (at least at time of writing), which makes me wonder if there are any general techniques for this.
One idea which was brought up (twice) is letting $G$ act on the cosets of $H$. Then if $G$ is simple, $G$ acts faithfully on the cosets, and so there is an injection \(G \hookrightarrow \mathfrak{S}_{[G:H]}.\) In particular, if the order of $G$ doesn’t divide $[G:H]!$, there can be no subgroup of order $H$.
I seem to remember seeing this idea on an exam from an undergraduate algebra midterm, but I can’t find the problem…
There’s a lot of results describing how the prime factorization of $G$ relates to the subgroups that it does have (I’ll discuss this some more in a footnote^{5}), but I’m still struggling to come up with more ways to show $G$ omits a subgroup of some order. If anyone has any ideas, I would love to hear about it, either at my mse question or in the comments here ^_^.

For completeness, we know each element of order $7$ is in exactly one $7$sylow subgroup. How many $7$sylow subgroups are there? Well, it’s $1$ mod $7$ and divides $\frac{168}{7} = 24$. It’s quick to see that the only choices are $1$ and $8$, but we know it cannot be $1$ (otherwise the $7$sylow subgroup would be normal, contradicting simplicity of $G$). So there are $8$ of them. Each contributes $6$ elements of order $7$ (since they each contain the identity), and we get $48$ elements total. ↩

Recall the normalizer of $P$ is the largest subgroup of $G$ in which $P$ is normal. ↩

As a little hint, consider the orbitstabilizer theorem.
As a big hint, let $G$ act on the set of sylow subgroups by conjugation, then consider orbitstabilizer. ↩

In fact, the mantra for proving things about finite $p$groups is “normalizers grow”. This mantra extends to finite nilpotent groups, since a finite nilpotent group is the direct product of its $p$sylow subgroups.
This doesn’t immediately seem helpful, since there are groups where some $p$sylow subgroup satisfies $N_G(P) = P$. Though since looking it up, it seems like this is actually a pretty safe bet. Self Normalizing Sylow Subgroups by Guralnick, Malle, and Navarro (available from the ams) shows that if $P$ is a $p$sylow subgroup for $p \gt 3$ then $P = N_G(P)$ implies $G$ is solvable!
Since our group is simple (and $7 \gt 3$), we could retroactively justify this idea. I like the Frattini Argument angle more, though, so that’s what I put in the main body of this post. ↩

The Sylow Theorems tell you that we can always find subgroups of prime power order, and more generally Hall’s Theorem tells us that if $G$ is solvable, then $G$ has a subgroup of order $k$ whenever $k$ and $\frac{G}{k}$ are coprime^{5}.
But we know that lots of groups are solvable! For instance, the famed FeitThompson Theorem, says that every group of odd order is solvable! Citing another famous result, we know by Burnside’s $p^aq^b$ Theorem that any group with only two prime divisors (that is, every group of order $p^aq^b$) is solvable too.
Moreover, it’s easy to see that every group of squarefree order is solvable (this is a cute exercise), and thus satisfies the converse of lagrange’s theorem (here we heavily use squarefreeness of $G$). So any group which omits a subgroup of some order must have a factor of $p^2$ for some prime $p$.
There may be odd order groups (or groups with only two prime divisors) which omit subgroups, but we see that they can’t omit many subgroups by Hall’s Theorem. By the remark about squarefree order groups, the simplest case is groups of order $p^2 q$, and indeed there are already groups of this order which omit a subgroup. We see this with $A_4$, of order $12 = 2^2 \cdot 3$, for instance, which famously has no subgroup of order $6$.
Groups which have subgroups of all possible orders are called CLT Groups for “Converse to Lagrange’s Theorem”. For more information about the relation between the prime factors of $G$ and the CLTness (or indeed the nonCLTness) of $G$, you should check out the (extremely readable!) thesis Groups Satisfying the Converse to Lagrange’s Theorem by Jonah N. Henry. You can find a copy here. ↩