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Consider $A_{5}$ group and let $a,b \in A_5 $ .

Find probability of the subgroup generated by a and b. $\langle a, b\rangle$ , be cyclic group .

I think if $a=1$ then we have 60 cyclic subgroup because $A_5$ has 60 elements, also 59 cyclic subgroup for $b=1$ so we have 60+59 cyclic subgroups and probability is $\frac{119}{60×60}$ is this true?

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    The number of cyclic subgroups is a lot smaller. For example, if $a=(12345)$, then $a$ and all its powers generate the same subgroup. There are six cyclic subgroups of order five, ten of order three and fifteen of order two. – Jyrki Lahtonen Mar 18 '21 at 14:56
  • @JyrkiLahtonen . I think we must only find a,b and this is not necessary to find this subgroups are isomorphic or not . –  Mar 18 '21 at 15:00
  • I think that this entry is relevant: https://math.stackexchange.com/questions/435952/probability-that-two-randomly-chosen-permutations-will-generate-s-n – Nicky Hekster Mar 18 '21 at 15:14
  • Related. – Shaun Mar 18 '21 at 15:50
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    What I was getting at is that if $a$ is 5-cycle then $\langle a,b\rangle$ is cyclic for five choices of $b$. If $a$ is a 3-cycle, then $\langle a,b\rangle$ is cyclic for three choices of $b$ etc. You compute the weighted average. I was not counting isomorphism types, I was counting distinct cyclic subgroups. – Jyrki Lahtonen Mar 18 '21 at 16:05
  • Anyway, the answer to the question is no that is not true. The correct answer, as given by C Monsour is 3/40. – Derek Holt Mar 20 '21 at 13:15

1 Answers1

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The orders of elements of $A_5$ are

$Order \quad Num\, Elements \\ \\ 1 \quad\quad\quad\quad 1 \\ 2 \quad\quad\quad\quad 15 \\ 3 \quad\quad\quad\quad 20 \\ 5 \quad\quad\quad\quad 24$

Since none of these numbers are composite, $<a,b>$ will be cyclic precisely when one of these elements is a power of the other. This is because if two elements generate a cyclic group and one is not a power of the other, then the cyclic group has composite order (in a group of prime order all elements are powers of any non-identity element, and the identity alone doesn't generate a group of prime order). But $A_5$ has no cyclic subgroups of composite order as it has no elements of composite order.

Since the problem does not state whether $a$ and $b$ are drawn with or without replacement, it's somewhat ambiguous, and it actually makes a rather large difference: $<a,b>$ is substantially more likely to be cyclic if drawn with replacement. However, in usual mathematical usage, one does not assume $a \ne b$ unless explicitly stated, so technically we should answer the problem as though these are drawn with replacement.

Therefore, out of $3600$ (i.e., $60\cdot 60$) cases, we have

$60$ cases of $a=1$ and $b$ any.

$30 = 2\cdot 15$ cases of $o(a)=2$ and $b\in <a>$

$60 = 3 \cdot 20$ cases of $o(a) = 3$ and $b\in <a>$

$120 = 5 \cdot 24$ cases of $o(a) = 5$ and $b\in <a>$

So the probability you are looking for is $270/3600$, or $7.5\%$

Also note that if you want the "draw without replacement answer" you can simply subtract the $60$ cases of $a=b$ from both the numerator and the denominator, to get $210/3540=7/118$, or just a hair over $5.9\%$.

C Monsour
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