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February 5th, 2017, 04:36 AM   #1
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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 04:47 AM.
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February 17th, 2017, 11:33 AM   #2
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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.
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