<PRE><B>Question 1167792</B>
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&#39;s pick an element $g \in A_4$. Let&#39;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&#39;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&#39;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&#39;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&#39;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&#39;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&#39;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&#39;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&#39;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&#39;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.</PRE>