In $S_7$, find the permutation $x$ that satisfies the condition $x^2 = (1,2)(3,4)$.

Question

In $S_7$, $X^2 = (1,2)(3,4)$.

What is the permutation $X$ that satisfies the given condition?

There are four of them. So asking about the permutation $X$ is a bit misleading. — Arthur, Aug 13 '15 at 13:05
You really should show what you have tried. I can find several elements of $S_7$ which would do for $X$ — Mark Bennet, Aug 13 '15 at 13:05
This is point $a)$ of a problem given at a exam in the University of Bucharest - Faculty of Mathematics and Informatics. No other information given.. — Nazare Andrei-Cristian, Aug 13 '15 at 13:16
I have found a solution to this problem stating that $x=(1324)(567)$ BUT $x^2=e$ so this doesn't satisfy the condition AND ALSO i must FIND the said $x$ -> resolving the said equation (somehow). — Nazare Andrei-Cristian, Aug 13 '15 at 13:30
no for $x=(1324)(567)$ you don't get $x^2=e$ because since $(1324)$ and $(567)$ are disjoint cycles, they commute and so $x^2=(1324)^2(567)^2=(12)(34)(576)$. Doing this, we find one solution of your equation and it is a 4-cycle, which is? — Scientifica, Aug 13 '15 at 13:44
No! We just calculated the square of $(1324)(567)$ and we found that it wasn't equal to $(12)(34)$ because there is an "additional" $(576)$ there. Where did $(576)$ come from? — Scientifica, Aug 13 '15 at 14:48
Please help me Scientifica. I have no clue what to do here or why to do it. I would guess that there's no such X that satisfies the condition. Please help! — Nazare Andrei-Cristian, Aug 13 '15 at 14:58
Aren't I helping? I'm actually trying to make you reach the answer. For why to do this, well every permutation has a unique cyclic decomposition, right? So to know what $x$ is let's study its cyclic decomposition. It has a cyclic decomposition, let's say $x=\sigma _1 \sigma _2\cdots \sigma _n$ for some $n\in\mathbb{Z}^+$. One of the first informations we would like to know is the length of the cycles. We know that the length of a cycle is its order and we also know that the order of each cycle $\sigma _i$ for $i\in{1,\cdots,n}$ divides the order of $x$ so knowing the order of $x$ is... — Scientifica, Aug 13 '15 at 15:08
... very important because then we will know all the possible lengths of the cycles and that's why @MarkBennet asked you to calculate it. So try to find it. Use the fact that in any group $G$ and for any $g\in G$ with $|g|$, the order of $g$, finite, and for any $a\in\mathbb{Z}^+$, we have $|g^a|=\frac{|g|}{(|g|,a)}$ where $(|g|,a)$ is the greatest common divisor of $|g|$ and $a$. If you can't calculate it, see my answer below (put the mouse on the blank between the two sentences you can see). — Scientifica, Aug 13 '15 at 15:12
$|x^2|=2$ means "The order of $x^2$ is $2$". What does this mean? (definition of the order of an element) — Scientifica, Aug 13 '15 at 15:16
If you are trying to do group theory and the theory of symmetric groups without understanding what the order of an element is, then you need to go back to your notes. If it is a language difficulty the order $r$ of an element $x$ is the least positive integer $r$ for which $x^r=$identity [for a group written multiplicatively] — Mark Bennet, Aug 13 '15 at 17:39
@MarkBennet and the definition isn't enough because he'll need some basic properties. It seems he knows what are orders since he didn't ask for it, but either he doesn't understand it or, and that's my fear, he doesn't know how to calculate products of cycles. — Scientifica, Aug 13 '15 at 21:39
don't know how to calculate products of cycles..i can only multiply permutations.. :( — Nazare Andrei-Cristian, Aug 14 '15 at 06:20
It's the same: the product of two permutations is their multiplication and it is their composition as applications. As Mark Bennet recommended you, you really should revise your course. I believe one should try to understand the most part of a course by revising the lesson and then do the exercises to complete the understanding and improve your skills. Doing exercises before understanding the lesson is a bad idea. Anyway, you know things in group theory, right? Mark Bennet reminded what the order of an element is. |g|=n means that $g^n=e$ and ... — Scientifica, Aug 14 '15 at 09:08
... for any positive integer $i$ strictly less than $n$, $g^i\neq e$. So to calculate the order of an element, the most straightforward way of doing it is to calculate its powers until you get $e$ (but not the one with less calculus. The use of Lagrange's Theorem is an example). The power that gives you $e$ for the first time is the order. Let $x$ be a solution of your equation. We have $x^2=(12)(34)$ and we want to find the order of $x^2$. Put $y=x^2$ to not confuse powers of $x$ and powers of $x^2$. $y\neq e$ right? Next step: calculate $y^2$... — Scientifica, Aug 14 '15 at 09:13
... $y^2=yy=(12)(34)(12)(34)$. Since $(12)$ and $(34)$ are disjoint cycles (recall what a cycle is and what do we mean by disjoint) then $(34)(12)=(12)(34)$ and so: $y^2=(12)(12)(34)(34)=(12)^2(34)^2$. We know that the order of a cycle is its length. The length of both $(12)$ and $(34)$ is $2$. Thus their order is $2$ and in particular their square is $e$ and so $y^2=e$ which means $|y|=|x^2|=2$. — Scientifica, Aug 14 '15 at 09:16
By the way, when you don't understand something you:1- Think about it. 2- Ask your teacher. 3- Search: Books, pdfs, videos and even Wikipedia can explain things well! 4- Ask in this forum. Asking is normal and no one should be ashamed to ask a question even in class. This link is a link to the last question I asked in this forum. See that it was about something I didn't understood in the course. (But in this forum, it's preferable when you ask help to an exercise to first... — Scientifica, Aug 14 '15 at 09:21
... show what you tried or at least what your thoughts were or simply "I don't know how to start"). — Scientifica, Aug 14 '15 at 09:22
Anyway back to your question, since $|x^2|=2$, using the fact that $|x^2|=\frac{|x|}{(|x|,2)}$ where $(|x|,2)$ is the greatest common divisor of $|x|$ and $2$, can you see why $|x|=4$? If not check my answer. Then check my previous comments about the cycle decomposition of $x$. Using what I told you, can you first at least tell me why the cycle decomposition of $x$ contains only $2$-cycles and $4$-cycles? — Scientifica, Aug 14 '15 at 09:26

Mark Bennet · Answer 1 · 2015-08-13T15:24:42.023

Hint: What would be the order of $X$?

As an alternative more hands-on approach, think what $X$ does to the element $1$ of the underlying set. After applying $X^2$ i.e. applying $X$ twice you get to $2$. After applying $X$ twice to $2$ you get to $1$. This is because $X^2$ contains the transposition $(1,2)$. So you have $$X: 1\to a\to 2\to b \to 1$$ where you don't know what $a$ and $b$ are.

What happens when you apply $X^2$ to $a$ and $b$? To $3$ and $4$? To $5, 6,$ and $7$?

score 2 · Answer 2 · answered Aug 20 '15 at 11:12

Write $X$ as product of disjoint cycles. Then you see what happens by squaring. Cycles of even length become two new cycles of half the length and cycles of odd length keep their length.

$X^2$ has no cycles of odd length so $X$ should also only have cycles of even length. We actually need a cycle of length 4 whose square is $X^2$. The two possibilities are $(1324)$, $(1423)$. Now we can multiply this by a cycle of length two whose square is the identity. There are the following possibilities: $(56)$, $(57)$, $(67)$. So in total we get $2\cdot 3=6$ possibilities for $X$.

score 1 · Answer 3 · answered Aug 13 '15 at 13:05

Edit: Since Mark Bennet asked to calculate the order as a hint, I'm hiding this part of my answer.

Note that $|X^2|=2$. Also: $|X^2|=\dfrac{|X|}{(|X|,2)}$ where (|X|,2) is the greatest common divisor of $|X|$ and $2$ and |X| is the order of $X$. Thus $|X|=2(|X|,2)$ and so $|X|$ is an even number. Thus $((|X|,2))=2$ and so $|X|=4$.

Try to find a 4-cycle which satisfies your equality.

In $S_7$, find the permutation $x$ that satisfies the condition $x^2 = (1,2)(3,4)$.

3 Answers3