Must a group of order $20$ have an element of order $10$?

Question

I'm trying to show that if $G$ is a group of order $20$, then it will have an element of order $10$.

Considering the Sylow $5$-subgroups, we have $n_5 \equiv 1 \mod 5$ and $n_5 | 4$, so $n_5=1$ and so we have one cyclic subgroup of order $5$ and thus $4$ elements of order $5$.

Similarly, $n_2 \equiv 1 \mod 2$ and $n_2|5$ so $n_2=1$ or $5$. If we have $n_2 = 5$ then we can have at most $5 \cdot 3 = 15$ elements of order $2$ in the group, which would occur if the Sylow $2$-subgroups are disjoint copies of $C_2 \times C_2$.

In this case, alongside the identity, we have accounted for all $1+4+15=20$ elements and we don't have an element of order $10$. Otherwise we have elements of order $10$ and $20$ and we are done.

So, I am wondering how (if?) we can rule out this one particular case, or if there is a better method for showing that there must be an element of order $10$.

Does this answer your question? Find four groups of order 20 not isomorphic to each other. — Anne Bauval, Feb 03 '23 at 14:50
@DietrichBurde I haven't encountered the classification of groups of order 20, in particular semidirect products and Galois groups - is it possible that there's a way to do this in a more elementary way? — Robin, Feb 03 '23 at 14:56
@Peter I'm not sure if I understand this notation, can you explan what you mean by $C_5 : C_4$? Thanks! — Robin, Feb 03 '23 at 14:56
Robin, no need to know the Galois group of $x^5-2$. It is the subgroup of $S_5$ generated by $(12345)$ and $(1243)$, so you can compute explicitly the order of elements! — Dietrich Burde, Feb 03 '23 at 14:58
@AnneBauval It possibly does but what I see in those answers is beyond what I've learned so far, is there a more elementary way to do this? Thanks. — Robin, Feb 03 '23 at 14:58
@DietrichBurde I see, thank you. I see that since $S_5$ does not have an element of order 10, neither does this. — Robin, Feb 03 '23 at 15:04
@DietrichBurde Everything is clear, thanks. If you want to write up an answer containing this then I'll accept it. — Robin, Feb 03 '23 at 15:51

calc ll · Answer 1 · 2023-02-03T17:14:45.643

4

Consider the holomorph $\rm hol(C_5)=C_5\rtimes \rm Aut(C_5)$. Thus it embeds in $S_5,$ which has no element of order $10.$

The is also the Frobenius group, or second Suzuki group $\rm Sz(2).$ It's the only counterexample.

edited Feb 03 '23 at 17:14

answered Feb 03 '23 at 15:42

calc ll

8,427

How do you know that it embeds into $S_5$? – citadel Feb 04 '23 at 10:23
That's a general fact about the holomorph. @Devo. $\rm hol(G)$ happens to embed in $S_{\lvert G\rvert}.$ – calc ll Feb 04 '23 at 10:25
1

See Derek Holt's two comments here https://math.stackexchange.com/a/4628060/1070376 @Devo – calc ll Feb 04 '23 at 10:31

score 3 · Accepted Answer · answered Feb 03 '23 at 15:54

The Frobenius group of order $20$, which arises as the Galois group of $X^5-2$ over $\Bbb Q$, is a subgroup of $S_5$, i.e., generated by $(12345)$ and $(1243)$. It is isomorphic to the semidirect product $C_5\rtimes C_4$. Since $S_5$ has no element of order $10$, it has no element of order $10$ either.

However, since all groups of order $20$ are supersolvable, they satisfy the converse of Lagrange: since $10$ divides $20$, there is a subgroup of order $10$ in all cases. For CLT-groups (converse Lagrange groups) see here:

Complete classification of the groups for which converse of Lagrange's Theorem holds

citadel · Answer 3 · 2023-02-05T07:54:51.863

If $G$ is abelian, then the answer is yes, because said $x$ and $y$ two elements of order $2$ and $5$ respectively (which both exist by Cauchy's theorem), the product $xy$ has order $10$. So, to look for possible counterexamples, we have to turn our attention to nonabelian groups, where semidirect products assist ($D_{10}$ is not a candidate, though, because by definition it has an element of order $10$). Now, $C_5$ acts only trivially by automorphisms on both $C_4$ and $C_2\times C_2$ (see e.g. here), a situation which leads to direct products of abelian groups, and hence abelian in turn. Thereofore, we are left to look for some truly semidirect product $(C_2\times C_2)\ltimes C_5$ or $C_4\ltimes C_5$. But the former would be equivalent to a homomorphism $\varphi\colon C_2\times C_2\to C_4$, which must be trivial as well, because a generator of $C_4$ can not be the image of elements of order $2$. So we are left to look at $C_4\ltimes C_5$ for a possible group without elements of order $10$, and both the other answers confirm that this is indeed the case.

For the downvoter: can you justify otherwise why to address ex nihilo abruptly $C_4\ltimes C_5$ as candidate? Looks like magics otherwise. — citadel, Feb 05 '23 at 07:00

Must a group of order $20$ have an element of order $10$?

3 Answers3

Linked