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Question 1167792: Let H = {(1),(13)(24)} in A4 .
(a) Show that H is not normal in A4.
(b) Show that (123)H = (243)H and (124)H = (132)H but that (123)(124)H 6= (243)(132)H . This proves that the group operation we defined on the set of (left) cosets G/H is not well defined unless we know that the subgroup H is normal.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $A_4$ be the alternating group on 4 elements, which consists of all even permutations of $\{1,2,3,4\}$.
The elements of $A_4$ are:
$A_4 = \{ (1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243) \}$
The order of $A_4$ is $|A_4| = 4!/2 = 12$.
Let $H = \{ (1), (13)(24) \}$.
The order of $H$ is $|H| = 2$.
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**(a) Show that H is not normal in A4.**
A subgroup $H$ is normal in a group $G$ (denoted $H \unlhd G$) if for all $g \in G$, $gHg^{-1} = H$. This means $gHg^{-1} \subseteq H$ is sufficient if $H$ is finite.
To show that $H$ is *not* normal in $A_4$, we need to find at least one element $g \in A_4$ such that $gHg^{-1} \neq H$. That is, $g h g^{-1} \notin H$ for some $h \in H$.
Let's pick an element $g \in A_4$. Let's try $g = (123)$.
The elements of $H$ are $h_1 = (1)$ and $h_2 = (13)(24)$.
1. For $h_1 = (1)$:
$g h_1 g^{-1} = (123)(1)(123)^{-1} = (123)(1)(132) = (1) \in H$. This is always true for the identity element.
2. For $h_2 = (13)(24)$:
$g h_2 g^{-1} = (123)(13)(24)(123)^{-1}$
First, find $g^{-1} = (123)^{-1} = (132)$.
So, $(123)(13)(24)(132)$.
Let's compute the product:
$1 \xrightarrow{(132)} 3 \xrightarrow{(13)} 1 \xrightarrow{(123)} 2$
$2 \xrightarrow{(132)} 1 \xrightarrow{(13)} 3 \xrightarrow{(123)} 1$
$3 \xrightarrow{(132)} 2 \xrightarrow{(13)} 2 \xrightarrow{(123)} 3$
$4 \xrightarrow{(132)} 4 \xrightarrow{(13)} 4 \xrightarrow{(123)} 4$
So, $(123)(13)(24)(132) = (12)(3)(4) = (12)$.
Now, check if $(12) \in H$.
$H = \{ (1), (13)(24) \}$.
Since $(12)$ is a 2-cycle, it is an odd permutation, thus $(12) \notin A_4$. This means $(123)(13)(24)(132)$ cannot be $(12)$.
Let's recompute the conjugation.
Conjugation of $(13)(24)$ by $(123)$:
$(123)(13)(24)(132)$
To conjugate a cycle $(a_1 a_2 \dots a_k)$ by a permutation $\sigma$, the result is $(\sigma(a_1) \sigma(a_2) \dots \sigma(a_k))$.
So, $(123)(13)(123)^{-1} = ((123)(1) (123)(3)) = (21)$.
And $(123)(24)(123)^{-1} = ((123)(2) (123)(4)) = (34)$.
So, $(123)(13)(24)(132) = (21)(34) = (12)(34)$.
Now, check if $(12)(34) \in H$.
$H = \{ (1), (13)(24) \}$.
Since $(12)(34) \neq (1)$ and $(12)(34) \neq (13)(24)$, it means $(12)(34) \notin H$.
Since we found an element $g = (123) \in A_4$ such that $gHg^{-1} = (123)H(123)^{-1} = \{ (1), (12)(34) \} \neq H$,
**H is not a normal subgroup of A4.**
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**(b) Show that (123)H = (243)H and (124)H = (132)H but that (123)(124)H $\neq$ (243)(132)H.**
**Recall left cosets:** $gH = \{gh \mid h \in H\}$.
**First, let's find the elements of the specified cosets:**
* **(123)H:**
$(123)(1) = (123)$
$(123)(13)(24) = (123 \cdot 13)(24) = (1)(23) = (23)(1)$ - no, (123)(13)=(1)(23)=(23).
Let's compute $(123)(13)(24)$:
$1 \xrightarrow{(13)(24)} 3 \xrightarrow{(123)} 1$
$2 \xrightarrow{(13)(24)} 4 \xrightarrow{(123)} 4$
$3 \xrightarrow{(13)(24)} 1 \xrightarrow{(123)} 2$
$4 \xrightarrow{(13)(24)} 2 \xrightarrow{(123)} 3$
So, $(123)(13)(24) = (243)$.
Thus, $(123)H = \{ (123), (243) \}$.
* **(243)H:**
$(243)(1) = (243)$
$(243)(13)(24)$:
$1 \xrightarrow{(13)(24)} 3 \xrightarrow{(243)} 2$
$2 \xrightarrow{(13)(24)} 4 \xrightarrow{(243)} 3$
$3 \xrightarrow{(13)(24)} 1 \xrightarrow{(243)} 4$
$4 \xrightarrow{(13)(24)} 2 \xrightarrow{(243)} 4$ - no, $4 \xrightarrow{(243)} 3 \xrightarrow{(13)(24)} 1 \xrightarrow{(243)} 4$ (mistake here)
Let's compute $(243)(13)(24)$:
$1 \xrightarrow{(13)(24)} 3 \xrightarrow{(243)} 2$
$2 \xrightarrow{(13)(24)} 4 \xrightarrow{(243)} 3$
$3 \xrightarrow{(13)(24)} 1 \xrightarrow{(243)} 4$
$4 \xrightarrow{(13)(24)} 2 \xrightarrow{(243)} 4$
Oh, $4 \xrightarrow{(13)(24)} 2 \xrightarrow{(243)} 4$. So $4$ maps to $4$.
So, $(243)(13)(24) = (123)$.
Thus, $(243)H = \{ (243), (123) \}$.
**Therefore, (123)H = (243)H.**
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* **(124)H:**
$(124)(1) = (124)$
$(124)(13)(24)$:
$1 \xrightarrow{(13)(24)} 3 \xrightarrow{(124)} 1$
$2 \xrightarrow{(13)(24)} 4 \xrightarrow{(124)} 1$
$3 \xrightarrow{(13)(24)} 1 \xrightarrow{(124)} 2$
$4 \xrightarrow{(13)(24)} 2 \xrightarrow{(124)} 4$
So, $(124)(13)(24) = (214)$ - no, $(214)$ is $1 \to 2 \to 4 \to 1$. $1 \to 3 \to 1$. $2 \to 4 \to 1$. $3 \to 1 \to 2$. $4 \to 2 \to 4$.
Let's recompute $(124)(13)(24)$:
$1 \xrightarrow{(13)} 3 \xrightarrow{(124)} 1$
$2 \xrightarrow{(24)} 4 \xrightarrow{(124)} 1$
$3 \xrightarrow{(13)} 1 \xrightarrow{(124)} 2$
$4 \xrightarrow{(24)} 2 \xrightarrow{(124)} 4$
So, $(124)(13)(24) = (21)(34)$. No, this is product of $(124)$ and $(13)(24)$.
$(124)(13)(24) = (1)(23)(4) = (23)$.
Let's compute $(124)(13)(24)$:
$1 \xrightarrow{(13)} 3 \xrightarrow{(124)} 1$ (1 goes to 1)
$2 \xrightarrow{(24)} 4 \xrightarrow{(124)} 1$ (2 goes to 1)
$3 \xrightarrow{(13)} 1 \xrightarrow{(124)} 2$ (3 goes to 2)
$4 \xrightarrow{(24)} 2 \xrightarrow{(124)} 4$ (4 goes to 4)
So $(124)(13)(24) = (213)$. No, $2 \to 1$, $3 \to 2$. This is $(123)$.
Let's use a permutation calculator or be extremely careful.
$(124)(13)(24)$
$1 \to 3 \to 1$
$2 \to 4 \to 1$
$3 \to 1 \to 2$
$4 \to 2 \to 4$
So $(124)(13)(24) = (1)(213)(4) = (123)$.
Thus, $(124)H = \{ (124), (123) \}$.
* **(132)H:**
$(132)(1) = (132)$
$(132)(13)(24)$:
$1 \xrightarrow{(13)(24)} 3 \xrightarrow{(132)} 2$
$2 \xrightarrow{(13)(24)} 4 \xrightarrow{(132)} 4$
$3 \xrightarrow{(13)(24)} 1 \xrightarrow{(132)} 3$
$4 \xrightarrow{(13)(24)} 2 \xrightarrow{(132)} 1$
So, $(132)(13)(24) = (124)$.
Thus, $(132)H = \{ (132), (124) \}$.
**Therefore, (124)H = (132)H.**
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**Now, let's check (123)(124)H $\neq$ (243)(132)H.**
**Left side: (123)(124)H**
First, calculate the product $(123)(124)$:
$1 \xrightarrow{(124)} 2 \xrightarrow{(123)} 1$
$2 \xrightarrow{(124)} 4 \xrightarrow{(123)} 4$
$3 \xrightarrow{(124)} 3 \xrightarrow{(123)} 2$
$4 \xrightarrow{(124)} 1 \xrightarrow{(123)} 3$
So, $(123)(124) = (243)$.
Now, calculate $(243)H$:
From earlier, $(243)H = \{ (243), (123) \}$.
So, **(123)(124)H = { (243), (123) }**.
**Right side: (243)(132)H**
First, calculate the product $(243)(132)$:
$1 \xrightarrow{(132)} 3 \xrightarrow{(243)} 2$
$2 \xrightarrow{(132)} 1 \xrightarrow{(243)} 4$
$3 \xrightarrow{(132)} 2 \xrightarrow{(243)} 3$
$4 \xrightarrow{(132)} 4 \xrightarrow{(243)} 1$
So, $(243)(132) = (124)$.
Now, calculate $(124)H$:
From earlier, $(124)H = \{ (124), (123) \}$.
So, **(243)(132)H = { (124), (123) }**.
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**Comparing the two results:**
(123)(124)H = { (243), (123) }
(243)(132)H = { (124), (123) }
Since $\{ (243), (123) \} \neq \{ (124), (123) \}$ (because $(243) \neq (124)$),
**Therefore, (123)(124)H $\neq$ (243)(132)H.**
This clearly demonstrates that the product of cosets is not well-defined if the subgroup is not normal. In general, if $aH=bH$ and $cH=dH$, it does not necessarily mean that $(ac)H = (bd)H$ unless $H$ is a normal subgroup.
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