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I am wondering about how to solve question 35 from chapter 5 (Permutation Groups) from the 10th edition of Gallian’s Abstract Algebra. The full question is as follows:

What is the smallest $n$ for which there is a solution in $S_n$ to the equation $x^5 = (12345)?$ Give an example of a solution. How many solutions are there for your $n$?

Okay, so we know that since $|x^5| = 5$, that $|x|=25$. From this we can hypothesize at the disjoint representation of $x$. Notice that in order for the least common multiple of the lengths to be 25, our only choice is to have a single 25-cycle in form $(a_1, a_2, … a_{24}, a_{25})$.

Okay, so $x^5$ is going to be the product in the form $(a_1, a_2, … a_{24}, a_{25})^5$, so notice that this will essentially have the effect of the below

$(a_1, a_2, … a_{24}, a_{25})^5$ = $(a_1, a_6, a_{11}, a_{16}, a_{21})(a_2, a_7, a_{12}, a_{17}, a_{22})(a_3, a_8, a_{13}, a_{18}, a_{23})(a_4, a_9, a_{14}, a_{19}, a_{24})(a_5, a_{10}, a_{15}, a_{20}, a_{25})$

I realized that you can pretty easily get an example $x$ such that one of these products are (1,2,3,4,5), for example (1, 6, 7, 8, 9, 2, 10, 11, 12, 13, 3, 14, 15, 16, 17, 4, 18, 19, 20, 21, 5, 22, 23, 24, 25), which turns out to be the answer given in the textbook. Furthermore, they say there are $20!$ possible answers, which probably follows pretty simply using some combinatorial logic and isn’t my big concern here.

But what I don’t get is why is this the answer? It seems wrong, because, this answer for $x$ doesn’t satisfy $x^5 = (12345)$, it just satisfies $x^5 = (12345)(a_{i_1}, …)(a_{i_6}, …)(a_{i_{11}}, …) (a_{i_{16}}, …)(a_{i_{21}}, …)$

Why is this the answer?

Thank you!

Shaun
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adam dhalla
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    I wonder, what edition of the textbook do you have? It might be useful to check if the question is still present in later editions… – Akiva Weinberger May 22 '22 at 07:41

1 Answers1

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I believe you are correct, and any solution that says you can do this is wrong. Consider $x$ as a product of disjoint cycles

$$\prod_{i=1}^m C_i\left(\prod_{i=1}^{m_5} {}_5C_i\right)$$

where each $C_i$ has order not divisible by $5$ and each ${}_5C_i$ has order $5k$ for some $k$.

So $$(12345)=x^5=\prod_{i=1}^m C_i^5\left(\prod_{i=1}^{m_5} {}_5C_i^5\right)$$

Each $C_i^5$ is still a cycle of the same order as $C_i$, disjoint from all the other cycles. So there must have been none of these in the first place.

So $$(12345)=x^5=\prod_{i=1}^{m_5} {}_5C_i^5$$

Each $5k$-cycle to the 5th power is a product of five disjoint $k$-cycles. (For $k=1$, identity $1$-cycles that collapse.) So the expression on the right is a product of disjoint cycles and there might be zero of them, or then at least five of them, but never just one. So it's impossible for the right side to be a single $5$-cycle.

Maybe if the question were worded differently: what is the smallest $n$ for which a disjoint cycle representation of $x^5$ can contain $(12345)$?

2'5 9'2
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  • Thank you! May I ask, why did you assume that all $C_i$ either have orders relatively prime to 5, 5, 25, … or the ensuing powers of 5? – adam dhalla May 22 '22 at 05:07
  • @DownstairsPanda Because there are different reasons for eliminating each block. It's maybe not the most efficient way. But I wanted to get to the point where only $25$-cycles are in play. – 2'5 9'2 May 22 '22 at 05:25
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    Sorry - am new to learning all this. I understand that, I just was wondering because some cycle of say, order 10 (not relatively prime to 5 and not exactly 5, 25, …) isn’t covered by these starting cases.

    Could it be reasonable to start off by saying since $x = |25|$, we know, the only choices are $5C$ and ${25}C$, where these are 5-cycles and 25-cycles respectively?

     – adam dhalla May 22 '22 at 06:33
  • @DownstairsPanda You are right. I wasn't thinking clearly. I'll delete this and maybe undelete it if I can fix it without much editing. – 2'5 9'2 May 22 '22 at 16:55
  • Oh, I can't delete an accepted answer. I will see if I can edit it now. – 2'5 9'2 May 22 '22 at 16:56
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    Alright, I edited the answer and I think it's not only correct now (hopefully), but also more concise. – 2'5 9'2 May 22 '22 at 17:02