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 $10$.

Answer 1

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

Answer 2

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.

Answer 3

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.