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While solving a DPP from my coaching, I came across the following question which bothered me:

Q13. If the element $a$ of a group $G$ is of order $n$, then choose the incorrect statement:

  1. $a^m = e$ if $n$ is a divisor of $m$.
  2. $n$ is a divisor of $m$ iff $a^m = e$.
  3. $n$ is not a divisor of $m$ if $a^m = e$.
  4. $n$ is not a divisor of $m$, iff $a^m = e$.

According to the answer key, the incorrect statement is (2), but that doesn’t seem right to me.

Here’s what I think:

  1. (1) is true by definition: if $\operatorname{ord}(a)=n$, then for any integer $k$, $$a^{kn} = e.$$

  2. (2) also seems true, since the minimality of $n$ implies $$a^m = e \quad\Longrightarrow\quad n \mid m,$$ and conversely if $n\mid m$ then $$a^m = (a^n)^{m/n} = e.$$

  3. (3) and (4) are clearly false, as they contradict this standard result: only multiples of $n$ can yield the identity.

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    Your analysis looks good to me. The book might be dealing incorrectly with $m=0$. – Ethan Bolker May 05 '25 at 13:12
  • This is four questions. – Shaun May 05 '25 at 13:59
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    @Shaun This is actually a single multiple-select (MSQ) type question, which is quite common in Indian entrance exams. It’s not four separate questions. –  May 05 '25 at 14:06
  • Hmm... even without knowing anything about order and divisibility logically at least 2 of the statements must be incorrect. 1 is a looser condition that must follow from 2, and 3 from 4. But 1 and 3 and 2 and 4 directly contradict each other. So either none are correct. Or 1 but not 2 and neither 3 nor 4 are correct. Or 1 and 2 but not 3 nor 4 are correct. Or 3 but not 4 and neither 1 nor 2 are correct. Or 3 and 4 are correct but neither 1 nor 2. That's without knowing anything. – fleablood May 05 '25 at 15:47
  • I think @EthanBolker interpretation is correct. $a^0 = e$. By my definition $n$ IS a divisor of $0$ (every integer divides $0$) but I can see how a bad text might assume $n$ is not a divisor of $0$. But even so question 3 and 4 should both be obviously false. – fleablood May 05 '25 at 15:51

1 Answers1

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That's right. $3$ and $4$ are false.

The relevant fact (and a nice exercise employing the division algorithm) is that $a^m=e$ implies that the order of $a$ divides $m.$ The converse is trivial.

suckling pig
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