# Solution clarification for the question: Prove that a group of order 33 contains an element of order 3

For the following question:

Prove that a group of order 33 contains an element of order 3.

From the student solution manual, the solution reads as follows:

Suppose $$|G|=33$$ and $$G$$ has no element of order 3. Then by Theorem 7.8 there is no element of order 33, and Lagrange implies that every nonidentity element has order 11. Count the number of subgroups of order 11 to obtain a contradiction.

My question about the solution is as follows:
By counting the elements of order 11, what contradiction am I suppose to obtain. Since my understanding is that by a corollary to Lagrange theorem (which is from a different text), that

In a finite group, the number of element of order $$d$$ is a multiple of $$\phi(d)$$. (Euler's Totient function)

Hence for $$d=11$$, we have $$\phi(11)=10.$$, and there are at most 30 of them. Lagrange theorem does not imply that a group of finite order must have elements of order 11. But since there are 30 elements of order 11. What contradiction am I suppose to be deriving from this conclusion?

• You wrote: "the order of an element order d is a multiple of ϕ(d)". It is logically incorrect. Commented Feb 21, 2022 at 7:38
• @HamidEnki thank you for pointing out my typo. I edited my post accordingly.
– Seth
Commented Feb 21, 2022 at 7:47
• By Lagrange the possible orders of elements are $1$, $3$, $11$, $33$: only the identity has order $1$ and $3$, $33$ are excluded: wo we have $32$ elements of order $11$; but these lie in subgroups of prime order $11$ which don't overlap, so that there are $10k$ elements of order $11$: contradiction. Commented Feb 21, 2022 at 7:55
• @ancientmathematician but can the group $G$ of order 33 have 30 elements of order 11 and not 32. I mean, if so, then won't a proof by contradiction be avoided?
– Seth
Commented Feb 21, 2022 at 8:10
• You don't need Sylow, Cauchy, whatever. Just as I said Lagrange plus the fact that the elements of order $11$ (which is prime) lie in subgroups of order $11$ which don't intersect except in the identity: so that the number of elements of order $11$ is $10k$. As $33\ne 10k+1$ you can't just have elements of order $11$ (plus the identity). Commented Feb 21, 2022 at 8:53

1. Assume $$G$$ has no element of order $$3$$. Then all of it's elements are either of order $$1$$ or $$11$$. But as you said correctly, the number of elements of order $$11$$ must be a multiple of $$\phi(11)$$ ( for proof see Gallian's contemporary abstract algebra). Then you have $$3\cdot \phi(11)=3\cdot 10 =30$$ elements of order $$11$$. And one element is the identity which has order $$1$$. This leaves you with a group of $$31$$ elements and not $$33$$. Contradiction.

Let me explain the flow of logic leading to such a conclusion.

step 0:- Apriori we cannot say how many elements are there of order 11 or 3 are in the group. All we have in hand is this theorem which says that the number of elements of order $$d$$ must be a multiple of $$\phi(d)$$.

Step 1:- Assume that there are no elements of order $$3$$.

step 2:- This leads to the conclusion that there must be atleast one element of order $$11$$. Of course all 33 elements cannot be identity duh!.

Step 3:- Now applying the theorem we have the number of elements of order $$11$$ are eithe $$10,20$$ or $$30$$. In all three cases you fall short of the number $$33$$ .

Step 4:- You conclude that your assumption was false and there must exist an element of order $$3$$.

Now let me explain how you cannot avoid proof by contradiction if you only know apriori Lagrange's theorem and that corollary. If you go about trying to classify the types of orders of elements are there in a group of order $$33$$ you end up with the following three possibilities :-

If there does exist an element of order $$11$$ then you have by arguing as Gallian does that you will end up with $$10,20$$ or $$30$$ elements of order $$11$$. Thus in this case you have to conclude that the remaining elements except the identity have to be of order $$3$$. Done !

If there does not exist an element of order $$11$$ then you must have $$32$$ elements of order $$3$$ in a group of order $$33$$ and hence you are done as you have proven there exists one(again all of them cannot be identity). Mind you that although this conclusion is absolutely mental, keeping in view bit more advanced theorems like Cauchy or Sylow and many other results ,you are not actually ending up violating what you have assumed to be true apriori. And thus you are done. This is obviously not very clean but because you are not breaking the logical flow and arriving to the conclusion you set out to prove , this suffices and proves the claim that there must exist one element of order $$3$$.

1. Use Cauchy's Theorem. If $$p$$ is a prime and $$p$$ divides $$|G|$$ then it has an element of order $$p$$.

3.Sylow's Theorem. If $$p^{r}\,,r\geq 1$$ divides $$|G|$$ then it has a subgroup of order $$p^{r}$$. From here you can conclude that the subgroup has an element of order $$3$$ as it is cyclic.

• Of course this is correct but it seems to me to be far too deep for such a trivial result! Commented Feb 21, 2022 at 8:54
• @Mr.GandalfSauron in Gallian's solution on the same question, he states that if $G$ has no elements of order 33, it has 30 elements of order 11 plus the identity element, then the remaining two elements must have order 3. The proof by contradictions is avoided entirely. I was initially confused as to whether $G$ can have order 11 if it has order 3. I was not aware of the result having to do with the euler totient function until I consulted Gallian's text.
– Seth
Commented Feb 21, 2022 at 8:55
• @SethMai you are not avoiding proof by contradiction. To conclude that $G$ has 30 elements of order $30$ you need to know the result that the elements of order $d$ occurs in multiples of $\phi(d)$. Thereafter you are eliminating the cases and if you assume that it has no elements of order $3$ , you then end up with a contradiction. Mind you there is no Structure Theorem here. The group may be abelian or non-abelian. So you cannot apriori predict that it has $30$ elements of order 11. I'll explain it in a bit more detail. Commented Feb 21, 2022 at 9:14
• @SethMai I have edited in more details to make it clear. Please check Commented Feb 21, 2022 at 9:29
• @Mr.GandalfSauron thaank you for the clarification. When I wrote out the proof by contradiction, the text did not even talk about the result concerning totient function. I only had to work with Lagrange theorem. So I wasn't sure if the case that it has elements of order 3, then, can it still have elements of order 11. Also, I wasn't sure when I see Cauchy's theorem, it has to do with the the existent of elements of order p as subgroups when p is prime....
– Seth
Commented Feb 21, 2022 at 9:51