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September 15th, 2014, 09:45 AM  #1 
Newbie Joined: May 2014 From: Singapore Posts: 11 Thanks: 0  Doubts from transposition, fixing and Identity
Dear All, Please see the attachment for the text i will be referring to. what does "I = τ1 · · · τm, where each τj is a transposition acting on {1, . . . , n}. Clearly, m != 1, thus m ≥ 2. Suppose, for the moment, that τm does not fix n". what does fixing n mean in a transposition? what are a.b and c??? I thought all transpositions started with 1? what does this whole sentence mean "It follows that we can now write I as a product of m transpositions in which the first transposition to be applied fixes n (this was proved under the assumption that τm(n) != n, and I is already in this form if τm(n) = n)." ? please help me understand lemma 1.4.3 with an example?? Thanks..... 
September 16th, 2014, 08:20 AM  #2  
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Quote:
I suggest getting more wellacquainted with $S_3$ which has 6 elements, and $S_4$ which has 24...the full symmetry groups get very big, fairly fast, and it becomes hard to write down everything explicitly for $n > 4$ (the multiplication table for $S_5$ has 14,400 entries). What we want to do is prove the PARITY (evenness/oddness) of a transposition decomposition of the identity is invariant. The "number" of transpositions it takes to do so is NOT invariant for example, in $S_3$: $I = (1\ 3)(1\ 3) = (1\ 3)(1\ 3)(1\ 2)(1\ 2) = (1\ 3)(1\ 2)(1\ 2)(1\ 2)(1\ 2)(1\ 3)$ so we have a way to write the identity as a product of 2,4 or 6 transpositions right there. Transpositions need not "start with a 1", in $S_4$ we have the following transpositions: (1 2), (1 3), (1 4), (2 3), (2 4), (3 4) of which only the first 3 involve 1. The first transposition only moves 1 and 2 (they swap places), so it fixes 3 and 4. Similarly (1 3) fixes 2 and 4, and so on. ********* Ok, so we have an induction argument to carry out. Our base case is $n = 2$. We assume that for some $n1$ (maybe just 2, we don't know yet) the theorem is true. We can thus regard our transpositions as elements of $S_{n1}$. Now all of these can also be regarded as elements of $S_n$ that fix $n$. Now we suppose we have some arbitrary product of transpositions in $S_n$ that equals the identity. Here's our strategy: Suppose that the last transposition fixes $n$. Then we have a SHORTER product: $\tau_1\cdots\tau_{m1}$ that fixes $n$ as well. If $\tau_{m1}$ ALSO fixes $n$, we keep going. If ALL of them fix $n$, then they all can be considered as lying in $S_{n1}$, and we can apply our induction hypothesis to conclude that $m$ is even. Of course, the assumption that $\tau_m(n) = n$ might not be TRUE. So here, we take a slightly different tack: we show we can rewrite our product with DIFFERENT transpositions, in such a way that $m$ isn't changed. So if $\tau_m(n) \neq n$ it must be that $\tau_m$ is of the form $(a\ n)$ for some number $a \in \{1,2,3,\dots,n1\}$. Next we look at the "nexttolast" transposition. There are 4 cases: 1. $\tau_{m1}(n) = n$ and $\tau_{m1}(a) = a$. This implies that $\tau_{m1}$ and $\tau_m$ are disjoint 2cycles, so they commute, and we can just switch the order. 2. 1. $\tau_{m1}(n) = n$ and $\tau_{m1}(a) = b$ <this is the same $a$ as we have in $\tau_m$. In this case, we have: $\tau_{m1}\tau_m = (a\ b)(a\ n) = (a\ n\ b) = (b\ n)(a\ b) = \sigma_{m1}\sigma_m$ Note we still have 2 transpositions at the end, but the product: $I = \tau_1\tau_2\cdots \tau_{m2}\sigma_{m1}\sigma_m$ has the same number of transpositions ($m$), but now the last one fixes $n$. 3. $\tau_{m1}(n) = a$. In this case $\tau_{m1} = \tau_m$, and we can either "cancel them" (reducing $m$ by 2, which doesn't change its parity), or replace them both with: $(a\ b)(a\ b)$ (where neither $a$ nor $b$ is $n$). 4. $\tau_{m1}(n) = b \neq a$. In this case, we have: $\tau_{m1}\tau_m = (b\ n)(a\ n) = (a\ b\ n) = (a\ n)(a\ b)$ so again, we can replace the last TWO transpositions with two more such that the last one fixes $n$. So in any case, we can arrange it so our last transposition fixes $n$, without changing $m$. We can keep doing this, until we've done this for every transposition except $\tau_1$. Now if $\tau_2\cdots\tau_m$ fixes $n$ since: $I = \tau_1(\tau_2\cdots\tau_m)$ it must be that $\tau_2\cdots\tau_m = (\tau_1)^{1} = \tau_1$. Since $\tau_2\cdots\tau_m$ fixes $n$ so does $\tau_1$, and thus the whole product fixes $n$we apply our induction hypothesis, and we are done.  

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doubts, fixing, identity, transposition 
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