
Abstract Algebra Abstract Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 5th, 2017, 03:36 AM  #1  
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  Determine the elements in the subgroups <(1234)> of S_4, and in <(1432),(24)> of S_4.
Hi, I have two exercise but I don't know how to proceed for the second one. I post what I have done for the first one: Exercise n.1: Determine the elements in the subgroup $\left \langle (1234) \right \rangle$ of $S_4$. Exercise n.2: The subgroup $\left \langle (1432),(24) \right \rangle$ of $S_4$ has order 8. Determine its elements and write each one as product of disjoint cycles. I have followed this definition: Quote:
Quote:
Resolution of the Exercise n.1: because of the iii) of the previous theorem I have considered only powers $n \ge 0$, and I had gone on by making attempts. for $n=0$: $(1234)^0 = (1)$ for $n=1$: $(1234)^1 = \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{matrix} \right ) \ne (1)$ for $n=2$: $(1234)^2 = \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{matrix} \right )^2 = \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{matrix} \right ) \circ \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{matrix} \right ) = \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{matrix} \right ) = (13)(24) \ne (1)$ for $n=3$: $(1234)^3 = (1234)^2(1234) = \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{matrix} \right ) \circ \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{matrix} \right ) = \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{matrix} \right ) = (1432) \ne (1)$ for $n=4$: $(1234)^4 = (1234)^3(1234) = \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{matrix} \right ) \circ \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{matrix} \right ) = \left ( \begin{matrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{matrix} \right ) = (1)$ we have just obtained again the identity element so, the elements in the subgroup $\left \langle (1234) \right \rangle$ are: $\left \langle (1234) \right \rangle = \left \{ (1), (1432), (13)(24), (1234)\right \}$. the order of the element $(1234)$: $o \left ( (1234) \right ) = \left  \left \langle (1234) \right \rangle \right  = 4$ i.e. there are 4 elements in $\left \langle (1234) \right \rangle$ but, how to resolve the second exercise? I think that it is similar to the first one, but, I don't know how to do it. Please, can you give me any suggestions? Many Thanks! Last edited by beesee; February 5th, 2017 at 03:47 AM.  
February 17th, 2017, 10:33 AM  #2 
Member Joined: Jan 2016 From: Athens, OH Posts: 51 Thanks: 27 
Let $H=<(1432),\,(24)>$. Now $(24)(1432)(24)^{1}=(24)(1432)(24)=(2341)=(1432)^{1}$ So $<1432>$ is normal in $H$. Let $a=(1432)\text{ and }b=(24)$. Then $a$ has order 4, $b$ has order 2, $<a>\cap<b>=<(1)>$ and $<a>$ is normal in H. So $H=<a><b>$ has order 8. Thus the elements of H are: $$\{(1),a,a^2,a^3,b,ab,a^2b,a^3b\}$$ I leave it to you to compute these elements as products of disjoint cycles. 

Tags 
<1234>, <1234>, <1432, 24>, determine, elements, subgroups 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
if sin(theta)= .1234 find (theta) in [4pi,pi]  mauro125  Algebra  3  February 22nd, 2014 04:57 PM 
Subgroups  gaussrelatz  Algebra  1  October 10th, 2012 11:30 PM 
Subgroups  DanielThrice  Abstract Algebra  1  November 25th, 2010 02:28 PM 
Subgroups Containing All Elements of Order n  Cider  Abstract Algebra  1  February 24th, 2010 07:18 AM 
no subgroups  bjh5138  Abstract Algebra  2  August 14th, 2008 12:03 PM 