 My Math Forum Determine the elements in the subgroups <(1234)> of S_4, and in <(1432),(24)> of S_4.
 User Name Remember Me? Password

 Abstract Algebra Abstract Algebra Math Forum

February 5th, 2017, 03:36 AM   #1
Senior Member

Joined: Jan 2013
From: Italy

Posts: 154
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:
 If $G$ is a group and $a \in G$, then $\left \langle a \right \rangle$ will denote the set of all integral powers of $a$. $\left \langle a \right \rangle = \left \{ a^n : n \in \mathbb{Z} \right \}$
and also this Theorem:
Quote:
 If $G$ is a group and $a \in G$, then $\left \langle a \right \rangle$, is a subgroup of $G$. Proof: We have to check the conditions for the a subgroup. i) $\left \langle a \right \rangle \ne \emptyset$ since $a \in \left \langle a \right \rangle$; ii) the set $\left \langle a \right \rangle$ is closed: if $a^m \in \left \langle a \right \rangle$ and $a^n \in \left \langle a \right \rangle$, then $a^m \cdot a^n = a^{m+n} \in \left \langle a \right \rangle$, since $m \in \mathbb{Z}, n \in \mathbb{Z}$ imply $m+n \in \mathbb{Z}$; iii) Finally, $\left \langle a \right \rangle$ contains the inverse of each of its elements, because if $a^m \in \left \langle a \right \rangle$, then $(a^m)^{-1} = a^{-m} \in \left \langle a \right \rangle$.

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: 93 Thanks: 48 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, $\cap=<(1)>$ and $$is normal in H. So H= has order 8. Thus the elements of H are:$$\{(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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mauro125 Algebra 3 February 22nd, 2014 04:57 PM gaussrelatz Algebra 1 October 10th, 2012 11:30 PM DanielThrice Abstract Algebra 1 November 25th, 2010 02:28 PM Cider Abstract Algebra 1 February 24th, 2010 07:18 AM bjh5138 Abstract Algebra 2 August 14th, 2008 12:03 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      