Romanian National Olympiad 1996 – Grade 12 – Problem 1

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Romanian National Olympiad 1996 – Grade 12 – Problem 1

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Source: Romanian National Olympiad 1996 – Grade 12 – Problem 1

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#1 h Apr 14, 2025, 10:12 AM

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Prove that a group $G$ in which exactly two elements other than the identity commute with each other is isomorphic to $\mathbb{Z}/3 \mathbb{Z}$ or $S_3.$

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#3 h Apr 14, 2025, 11:25 AM

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For any $1 \ne x \in G,$ the elements $x,x^2,x^3$ commute with each other. Since there are only two non-identity elements in $G$ that commute with each other, we must have $x^2=1$ or $x^3=1.$ Thus $|G|=2^m3^n,$ for some integers $m,n \ge 0.$ If $m \ge 2$ or $n \ge 2,$ then $G$ will have a subgroup of order $4$ or $9,$ which are abelian, contradiction. So $m,n \le 1$ and hence, since $|G| \ge 3,$ we get that $|G|=3$ or $|G|=6.$ If $|G|=3,$ then $G \cong \mathbb{Z}_3$ and if $|G|=6,$ then $G \cong S_3,$ the only non-abelian group of order $6.$ It is easy to see that both $\mathbb{Z}_3$ and $S_3$ satisfy the condition given in the problem.

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#4 h Apr 14, 2025, 10:20 PM

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what is S3

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#5 h Apr 15, 2025, 8:49 AM

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The symmetric group on $3$ elements.

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#6 h Apr 16, 2025, 12:57 PM

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We are given that in the group $G$ , there are exactly two non-identity elements that commute with each other. That is, there exist distinct elements $a, b \in G$ , $a \neq e$ , $b \neq e$ , such that $ab = ba$ , and all other pairs of non-identity elements in $G$ do not commute. Our goal is to determine all such groups $G$ up to isomorphism.

Step 1: The group cannot be abelian

Suppose $G$ were abelian. Then every pair of elements in $G$ commutes. In particular, all non-identity elements would commute with each other, violating the condition that only two non-identity elements commute. Therefore, $G$ must be a non-abelian group.

Step 2: Consider the number of commuting elements

We are told that exactly two non-identity elements commute. Denote these two elements by $x$ and $y$ . Then $xy = yx$ . We may assume $x \neq y$ and $x, y \neq e$ .

Observation: Any subgroup generated by $x$ and $y$ must be abelian, since $x$ and $y$ commute. But since no other pair of non-identity elements in $G$ commutes, this subgroup cannot contain more than three elements (identity and $x$ , $y$ ). Otherwise, we would get additional pairs of commuting non-identity elements.

Thus, the subgroup $\langle x, y \rangle$ must be of order 3. The only group of order 3 is the cyclic group $\mathbb{Z}/3\mathbb{Z}$ .

Step 3: Consider possible orders of $G$

Let us now determine the possible orders of $G$ . Since $G$ is finite (implied by the condition), suppose $|G| = n$ . Let us consider the possible prime divisors of $n$ .

By Cauchy's Theorem, for any prime $p$ dividing $|G|$ , there exists an element of order $p$ , and hence a cyclic subgroup of order $p$ .

Important point: Any cyclic subgroup of order $p > 3$ contains more than two non-identity elements (specifically, $p - 1$ of them), and these all commute with each other. That would contradict the assumption that only two non-identity elements in $G$ commute.

Hence, the only primes that can divide $|G|$ are $2$ and $3$ . That is,
$|G| = 2^m \cdot 3^n \quad \text{for some integers } m, n \geq 0.$
Now suppose $m \geq 2$ . Then $G$ would have a subgroup of order $4$ by Cauchy's Theorem and the Sylow theorems. All groups of order $4$ are abelian, so that subgroup would contain more than two non-identity elements that commute with each other, again violating the hypothesis.

Similarly, if $n \geq 2$ , then $G$ would have a subgroup of order $9$ , which is abelian (since all groups of order $9$ are abelian). Again, we would get too many commuting non-identity elements.

Therefore, we must have $m \leq 1$ and $n \leq 1$ . So the possible orders of $G$ are:
$1, 2, 3, 6.$
We can eliminate some of these immediately:

$|G| = 1$ : trivial group, no non-identity elements --- ruled out.

$|G| = 2$ : cyclic of order 2 --- abelian, so all elements commute --- ruled out.

$|G| = 3$ : cyclic of order 3. This group has two non-identity elements, and they commute (since it is abelian). This exactly matches the condition! So $\mathbb{Z}/3\mathbb{Z}$ is a valid example.

$|G| = 6$ : There are exactly two groups of order 6 up to isomorphism: the cyclic group $\mathbb{Z}_6$ , which is abelian (ruled out), and the symmetric group $S_3$ , which is non-abelian. We now check whether $S_3$ satisfies the condition.

Step 4: Check $S_3$

The group $S_3$ has six elements:
$S_3 = \{e, (12), (13), (23), (123), (132)\}.$
Among the non-identity elements, only the two 3-cycles $(123)$ and $(132)$ commute with each other:
$(123)(132) = (132)(123) = e.$ All transpositions (like $(12)$ and $(13)$ ) do not commute with each other, and neither do they commute with the 3-cycles. Thus, in $S_3$ , the only pair of non-identity elements that commute is $\{(123), (132)\}$ .

Therefore, $S_3$ satisfies the condition.

Conclusion

The only groups in which exactly two non-identity elements commute with each other are:
$\mathbb{Z}_3 \quad \text{and} \quad S_3.$
Important note:
These are not the only groups satisfying the given condition but the only one up to isomorhism.

This post has been edited 1 time. Last edited by MeKnowsNothing, Apr 16, 2025, 12:59 PM

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