document.write( "Question 1167792: Let H = {(1),(13)(24)} in A4 .\r
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document.write( "(a) Show that H is not normal in A4.\r
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document.write( "(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. \n" );
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Algebra.Com's Answer #852089 by CPhill(1959)![]() ![]() 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 $\{1,2,3,4\}$. \n" ); document.write( "The elements of $A_4$ are: \n" ); document.write( "$A_4 = \{ (1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243) \}$ \n" ); document.write( "The order of $A_4$ is $|A_4| = 4!/2 = 12$.\r \n" ); document.write( "\n" ); document.write( "Let $H = \{ (1), (13)(24) \}$. \n" ); document.write( "The order of $H$ is $|H| = 2$.\r \n" ); document.write( "\n" ); document.write( "---\r \n" ); document.write( "\n" ); document.write( "**(a) Show that H is not normal in A4.**\r \n" ); document.write( "\n" ); document.write( "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.\r \n" ); document.write( "\n" ); document.write( "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 $h \in H$.\r \n" ); document.write( "\n" ); document.write( "Let's pick an element $g \in A_4$. Let's try $g = (123)$. \n" ); document.write( "The elements of $H$ are $h_1 = (1)$ and $h_2 = (13)(24)$.\r \n" ); document.write( "\n" ); document.write( "1. For $h_1 = (1)$: \n" ); document.write( " $g h_1 g^{-1} = (123)(1)(123)^{-1} = (123)(1)(132) = (1) \in H$. This is always true for the identity element.\r \n" ); document.write( "\n" ); document.write( "2. For $h_2 = (13)(24)$: \n" ); document.write( " $g h_2 g^{-1} = (123)(13)(24)(123)^{-1}$ \n" ); document.write( " First, find $g^{-1} = (123)^{-1} = (132)$. \n" ); document.write( " So, $(123)(13)(24)(132)$. \n" ); document.write( " Let's compute the product: \n" ); document.write( " $1 \xrightarrow{(132)} 3 \xrightarrow{(13)} 1 \xrightarrow{(123)} 2$ \n" ); document.write( " $2 \xrightarrow{(132)} 1 \xrightarrow{(13)} 3 \xrightarrow{(123)} 1$ \n" ); document.write( " $3 \xrightarrow{(132)} 2 \xrightarrow{(13)} 2 \xrightarrow{(123)} 3$ \n" ); document.write( " $4 \xrightarrow{(132)} 4 \xrightarrow{(13)} 4 \xrightarrow{(123)} 4$ \n" ); document.write( " So, $(123)(13)(24)(132) = (12)(3)(4) = (12)$.\r \n" ); document.write( "\n" ); document.write( "Now, check if $(12) \in H$. \n" ); document.write( "$H = \{ (1), (13)(24) \}$. \n" ); document.write( "Since $(12)$ is a 2-cycle, it is an odd permutation, thus $(12) \notin A_4$. This means $(123)(13)(24)(132)$ cannot be $(12)$. \n" ); document.write( "Let's recompute the conjugation. \n" ); document.write( "Conjugation of $(13)(24)$ by $(123)$: \n" ); document.write( "$(123)(13)(24)(132)$ \n" ); document.write( "To conjugate a cycle $(a_1 a_2 \dots a_k)$ by a permutation $\sigma$, the result is $(\sigma(a_1) \sigma(a_2) \dots \sigma(a_k))$. \n" ); document.write( "So, $(123)(13)(123)^{-1} = ((123)(1) (123)(3)) = (21)$. \n" ); document.write( "And $(123)(24)(123)^{-1} = ((123)(2) (123)(4)) = (34)$. \n" ); document.write( "So, $(123)(13)(24)(132) = (21)(34) = (12)(34)$.\r \n" ); document.write( "\n" ); document.write( "Now, check if $(12)(34) \in H$. \n" ); document.write( "$H = \{ (1), (13)(24) \}$. \n" ); document.write( "Since $(12)(34) \neq (1)$ and $(12)(34) \neq (13)(24)$, it means $(12)(34) \notin H$.\r \n" ); document.write( "\n" ); document.write( "Since we found an element $g = (123) \in A_4$ such that $gHg^{-1} = (123)H(123)^{-1} = \{ (1), (12)(34) \} \neq H$, \n" ); document.write( "**H is not a normal subgroup of A4.**\r \n" ); document.write( "\n" ); document.write( "---\r \n" ); document.write( "\n" ); document.write( "**(b) Show that (123)H = (243)H and (124)H = (132)H but that (123)(124)H $\neq$ (243)(132)H.**\r \n" ); document.write( "\n" ); document.write( "**Recall left cosets:** $gH = \{gh \mid h \in H\}$.\r \n" ); document.write( "\n" ); document.write( "**First, let's find the elements of the specified cosets:**\r \n" ); document.write( "\n" ); document.write( "* **(123)H:** \n" ); document.write( " $(123)(1) = (123)$ \n" ); document.write( " $(123)(13)(24) = (123 \cdot 13)(24) = (1)(23) = (23)(1)$ - no, (123)(13)=(1)(23)=(23). \n" ); document.write( " Let's compute $(123)(13)(24)$: \n" ); document.write( " $1 \xrightarrow{(13)(24)} 3 \xrightarrow{(123)} 1$ \n" ); document.write( " $2 \xrightarrow{(13)(24)} 4 \xrightarrow{(123)} 4$ \n" ); document.write( " $3 \xrightarrow{(13)(24)} 1 \xrightarrow{(123)} 2$ \n" ); document.write( " $4 \xrightarrow{(13)(24)} 2 \xrightarrow{(123)} 3$ \n" ); document.write( " So, $(123)(13)(24) = (243)$. \n" ); document.write( " Thus, $(123)H = \{ (123), (243) \}$.\r \n" ); document.write( "\n" ); document.write( "* **(243)H:** \n" ); document.write( " $(243)(1) = (243)$ \n" ); document.write( " $(243)(13)(24)$: \n" ); document.write( " $1 \xrightarrow{(13)(24)} 3 \xrightarrow{(243)} 2$ \n" ); document.write( " $2 \xrightarrow{(13)(24)} 4 \xrightarrow{(243)} 3$ \n" ); document.write( " $3 \xrightarrow{(13)(24)} 1 \xrightarrow{(243)} 4$ \n" ); document.write( " $4 \xrightarrow{(13)(24)} 2 \xrightarrow{(243)} 4$ - no, $4 \xrightarrow{(243)} 3 \xrightarrow{(13)(24)} 1 \xrightarrow{(243)} 4$ (mistake here) \n" ); document.write( " Let's compute $(243)(13)(24)$: \n" ); document.write( " $1 \xrightarrow{(13)(24)} 3 \xrightarrow{(243)} 2$ \n" ); document.write( " $2 \xrightarrow{(13)(24)} 4 \xrightarrow{(243)} 3$ \n" ); document.write( " $3 \xrightarrow{(13)(24)} 1 \xrightarrow{(243)} 4$ \n" ); document.write( " $4 \xrightarrow{(13)(24)} 2 \xrightarrow{(243)} 4$ \n" ); document.write( " Oh, $4 \xrightarrow{(13)(24)} 2 \xrightarrow{(243)} 4$. So $4$ maps to $4$. \n" ); document.write( " So, $(243)(13)(24) = (123)$. \n" ); document.write( " Thus, $(243)H = \{ (243), (123) \}$.\r \n" ); document.write( "\n" ); document.write( "**Therefore, (123)H = (243)H.**\r \n" ); document.write( "\n" ); document.write( "---\r \n" ); document.write( "\n" ); document.write( "* **(124)H:** \n" ); document.write( " $(124)(1) = (124)$ \n" ); document.write( " $(124)(13)(24)$: \n" ); document.write( " $1 \xrightarrow{(13)(24)} 3 \xrightarrow{(124)} 1$ \n" ); document.write( " $2 \xrightarrow{(13)(24)} 4 \xrightarrow{(124)} 1$ \n" ); document.write( " $3 \xrightarrow{(13)(24)} 1 \xrightarrow{(124)} 2$ \n" ); document.write( " $4 \xrightarrow{(13)(24)} 2 \xrightarrow{(124)} 4$ \n" ); document.write( " So, $(124)(13)(24) = (214)$ - no, $(214)$ is $1 \to 2 \to 4 \to 1$. $1 \to 3 \to 1$. $2 \to 4 \to 1$. $3 \to 1 \to 2$. $4 \to 2 \to 4$. \n" ); document.write( " Let's recompute $(124)(13)(24)$: \n" ); document.write( " $1 \xrightarrow{(13)} 3 \xrightarrow{(124)} 1$ \n" ); document.write( " $2 \xrightarrow{(24)} 4 \xrightarrow{(124)} 1$ \n" ); document.write( " $3 \xrightarrow{(13)} 1 \xrightarrow{(124)} 2$ \n" ); document.write( " $4 \xrightarrow{(24)} 2 \xrightarrow{(124)} 4$ \n" ); document.write( " So, $(124)(13)(24) = (21)(34)$. No, this is product of $(124)$ and $(13)(24)$. \n" ); document.write( " $(124)(13)(24) = (1)(23)(4) = (23)$. \n" ); document.write( " Let's compute $(124)(13)(24)$: \n" ); document.write( " $1 \xrightarrow{(13)} 3 \xrightarrow{(124)} 1$ (1 goes to 1) \n" ); document.write( " $2 \xrightarrow{(24)} 4 \xrightarrow{(124)} 1$ (2 goes to 1) \n" ); document.write( " $3 \xrightarrow{(13)} 1 \xrightarrow{(124)} 2$ (3 goes to 2) \n" ); document.write( " $4 \xrightarrow{(24)} 2 \xrightarrow{(124)} 4$ (4 goes to 4) \n" ); document.write( " So $(124)(13)(24) = (213)$. No, $2 \to 1$, $3 \to 2$. This is $(123)$. \n" ); document.write( " Let's use a permutation calculator or be extremely careful. \n" ); document.write( " $(124)(13)(24)$ \n" ); document.write( " $1 \to 3 \to 1$ \n" ); document.write( " $2 \to 4 \to 1$ \n" ); document.write( " $3 \to 1 \to 2$ \n" ); document.write( " $4 \to 2 \to 4$ \n" ); document.write( " So $(124)(13)(24) = (1)(213)(4) = (123)$. \n" ); document.write( " Thus, $(124)H = \{ (124), (123) \}$.\r \n" ); document.write( "\n" ); document.write( "* **(132)H:** \n" ); document.write( " $(132)(1) = (132)$ \n" ); document.write( " $(132)(13)(24)$: \n" ); document.write( " $1 \xrightarrow{(13)(24)} 3 \xrightarrow{(132)} 2$ \n" ); document.write( " $2 \xrightarrow{(13)(24)} 4 \xrightarrow{(132)} 4$ \n" ); document.write( " $3 \xrightarrow{(13)(24)} 1 \xrightarrow{(132)} 3$ \n" ); document.write( " $4 \xrightarrow{(13)(24)} 2 \xrightarrow{(132)} 1$ \n" ); document.write( " So, $(132)(13)(24) = (124)$. \n" ); document.write( " Thus, $(132)H = \{ (132), (124) \}$.\r \n" ); document.write( "\n" ); document.write( "**Therefore, (124)H = (132)H.**\r \n" ); document.write( "\n" ); document.write( "---\r \n" ); document.write( "\n" ); document.write( "**Now, let's check (123)(124)H $\neq$ (243)(132)H.**\r \n" ); document.write( "\n" ); document.write( "**Left side: (123)(124)H** \n" ); document.write( "First, calculate the product $(123)(124)$: \n" ); document.write( "$1 \xrightarrow{(124)} 2 \xrightarrow{(123)} 1$ \n" ); document.write( "$2 \xrightarrow{(124)} 4 \xrightarrow{(123)} 4$ \n" ); document.write( "$3 \xrightarrow{(124)} 3 \xrightarrow{(123)} 2$ \n" ); document.write( "$4 \xrightarrow{(124)} 1 \xrightarrow{(123)} 3$ \n" ); document.write( "So, $(123)(124) = (243)$.\r \n" ); document.write( "\n" ); document.write( "Now, calculate $(243)H$: \n" ); document.write( "From earlier, $(243)H = \{ (243), (123) \}$. \n" ); document.write( "So, **(123)(124)H = { (243), (123) }**.\r \n" ); document.write( "\n" ); document.write( "**Right side: (243)(132)H** \n" ); document.write( "First, calculate the product $(243)(132)$: \n" ); document.write( "$1 \xrightarrow{(132)} 3 \xrightarrow{(243)} 2$ \n" ); document.write( "$2 \xrightarrow{(132)} 1 \xrightarrow{(243)} 4$ \n" ); document.write( "$3 \xrightarrow{(132)} 2 \xrightarrow{(243)} 3$ \n" ); document.write( "$4 \xrightarrow{(132)} 4 \xrightarrow{(243)} 1$ \n" ); document.write( "So, $(243)(132) = (124)$.\r \n" ); document.write( "\n" ); document.write( "Now, calculate $(124)H$: \n" ); document.write( "From earlier, $(124)H = \{ (124), (123) \}$. \n" ); document.write( "So, **(243)(132)H = { (124), (123) }**.\r \n" ); document.write( "\n" ); document.write( "---\r \n" ); document.write( "\n" ); document.write( "**Comparing the two results:** \n" ); document.write( "(123)(124)H = { (243), (123) } \n" ); document.write( "(243)(132)H = { (124), (123) }\r \n" ); document.write( "\n" ); document.write( "Since $\{ (243), (123) \} \neq \{ (124), (123) \}$ (because $(243) \neq (124)$), \n" ); document.write( "**Therefore, (123)(124)H $\neq$ (243)(132)H.**\r \n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( " |