# On finite groups

Let $(G,*)$ is a finite group, whose order is $n$. Also let $p|n$ and $p\in P$. Can you bring an example of such group which doesn't have an element of order $p$.

Sincerely,

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Please see the FAQ concerning signatures in your posts. Thank you. – Arturo Magidin Nov 6 '11 at 23:06
OK! Will take it into account next times! ;) – Tigran Hakobyan Nov 7 '11 at 18:42

As Ragib said, Cauchy's Theorem assures that if $p$ is a prime divisor of the order of the finite group $G$, then there is an element $g\in G$ of order $p$, hence at least $p-1$ of them. One may note that the proof of Cauchy Theorem's relies on a counting elements argument, thus it is not constructive.

In the comments, the questioner extended his question to elements of power prime order. Sylow's Theorems say that if $p^k$ is the highest power of the prime $p$ dividing the order of $G$, then there exists a subgroup $P<G$ with $|P|=p^k$ (such subgroups are called $p$-sylows). One can then apply the theory of $p$-grous to conclude that $G$ has subgroups of order any power $p^\ell$ with $\ell\leq k$. Again, the proof is not constructive.

Now, in order to decide about the existence of elements of order $p^k$ is enough to consider just one $p$-sylow and decide whether it needs to be cyclic (because by the Sylow's Theorems again all $p$-sylows are conjugated inside $G$, hence isomorphic one to another).

To answer (in the negative) to the general question it is enough to observe that in the permutation group on $10$ elements, a $5$-sylow has $25$ elements and one such is the subgroup $$P=\langle(1\ 2\ 3\ 4\ 5)\rangle\times\langle(6\ 7\ 8\ 9\ 10)\rangle$$ which is visibly non-cyclic.

Of course, one can come up with many more counterexamples, some of which much simpler than the one described above.

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One will not be able to find an example, because Cauchy's Theorem ensures there certainly is such an element.

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Thank you for your response. And what can we say, if we don't know that $p$ is a prime number? – Tigran Hakobyan Nov 6 '11 at 13:21
Nothing. There are examples where $n \mid o(G)$ but $n$ is not prime, and there exist no subgroups of order $n$, let alone an element with order $n.$ For example, the alternating group $A_4$ has no subgroups of order $6$, because subgroups with index $2$ must be normal and $A_4$ is a simple group. There exists elementary arguments to show $A_4$ has no subgroups of order $6$ as well. – Ragib Zaman Nov 6 '11 at 13:25
And if $p=q^k, q\in P$, then by Sylow's theorem there is a subgroup of order $p$. Is there an example of such group which doesn't have an element of order $p$? – Tigran Hakobyan Nov 6 '11 at 13:27
Could you clarify what $p$ means this time? If it is the order of the group, then yes there are groups which have no elements of order $p.$ Indeed, the only groups which have an element whose order is equal to the order of the group, are cyclic groups generated by that element. – Ragib Zaman Nov 6 '11 at 13:31
@RagibZaman You are correct that there is no subgroup of order $6$ in $A_4$ but your argument is not valid since $A_4$ is not simple. The smallest non-Abelian simple group is $A_5$. – Sebastian Schoennenbeck Nov 6 '11 at 14:50

Proposition: A positive integer n > 1 is prime if and only if every finite group whose order is divisible by n contains an element of order n.

Proof: If n is prime, then Cauchy's theorem guarantees an element of order n in any group whose order is divisible by n. If n is not prime, then n must be divisible by at least two primes, so that n = dpq. We consider three cases: (1) p = q, (2) p divides q−1, and (3) p < q. In the first two cases we find a group of order exactly n with no element of order n, but in the third case we merely find a group whose order is divisible by n with no element of order n. (1) If n is divisible by the square of a prime p, then n = dp2 and the group Cd×Cp×Cp has order divisible by (and exactly equal to) n, but contains no elements of order n. (2) If n = dpq is divisible by a product of distinct primes p, q such that p divides q−1 but such that neither p nor q divides d, then the group Cd×(Cp⋉Cq) has order divisible by (and exactly equal to) n, but contains no element of order n. (3) More generally, if n = dpq where p < q are primes that do not divide d, then the group Cd×Sym(q) has order divisible by (but not equal to) n and has no element of order n.

In the third case, if d = 1 but p does not divide q−1, then every group of order (exactly equal to) n = dpq has an element of order n, so the larger group is necessary in case d = 1 at least (and so necessary for this simple proof that "ignores" d).

The groups: I believe the properties of the example groups used are well known, but I state them explicitly here:

• Cn is the (unique up to isomorphism) cyclic group of order n. It has elements of order k exactly when k divides n.

• If p, q are primes with p dividing q−1 then Cp⋉Cq is the (unique up to isomorphism) non-abelian group of order pq. It has a unique Sylow q-subgroup of order q, and it has q Sylow p-subgroups, so that every element is either the identity, an element of order q, or an element of order p.

• Sym( q ) is the symmetric group on q points and every element of order q is a q-cycle. It has no elements of order dq for any d ≥ 2 (since their dth power is a q-cycle, they themselves must be a q-cycle, a contradiction), in particular, no elements of order pq.

• Cd × G has elements of order k exactly when there are positive integers i, j such that k = ij, i divides d, i is coprime to j, and j is the order of an element of G.

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