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text: 'Here are some additional examples of lattices. $\\newcommand{\\nsubg}{\\mathcal N \\mbox{-} Sub~G}$\n\nA familiar example\n---------------------------------\n\nConsider the following lattice.\n\n\n\nDoes this lattice look at all familiar to you? From some other area of mathematics, perhaps?\n\n%%hidden(Reveal the truth):\n\nIn fact, this lattice corresponds to boolean logic, as can be seen when we replace b with true and a with false in the following "truth table".\n\n\n\n%%comment:\n\nLatex source:\n\n\\begin{tabular} {| c | c | c | c |}\n \\hline\n $x$ & $y$ & $x \\vee y$ & $x \\wedge y$ \\\\ \\hline\n $a$ & $a$ & $a$ & $a$ \\\\ \\hline\n $a$ & $b$ & $b$ & $a$ \\\\ \\hline\n $b$ & $a$ & $b$ & $a$ \\\\ \\hline\n $b$ & $b$ & $b$ & $b$ \\\\ \\hline\n\\end{tabular}\n\n%%\n\n%%\n\n\nNormal subgroups\n---------------------------------\n\nLet $G$ be a group, and let $\\nsubg$ be the set of all [4h6 normal subgroups] of $G$. Then $\\langle \\nsubg, \\subseteq \\rangle$ is a lattice where for $H, K \\in \\nsubg$, $H \\wedge K = H \\cap K$, and $H \\vee K = HK = \\{ hk \\mid h \\in H, k \\in K \\}$.\n\n%%hidden(Proof):\n\nLet $H,K \\in \\nsubg$. Then $H \\wedge K = H \\cap K$. We first note that $H \\cap K$ is a [576 subgroup] of $G$. For let $a,b \\in H \\cap K$. Since $H$ is a group, $a \\in H$, and $b \\in H$, we have $ab \\in H$. Likewise, $ab \\in K$. Combining these, we have $ab \\in H \\cap K$, and so $H \\cap K$ is satisfies the closure requirement for subgroups. Since $H$ and $K$ are groups, $a \\in H$, and $a \\in K$, we have $a^{-1} \\in H$ and $a^{-1} \\in K$. Hence, $a^{-1} \\in H \\cap K$, and so $H \\cap K$ satisfies the inverses requirement for subgroups. Since $H$ and $K$ are subgroups of $G$, we have $e \\in H$ and $e \\in K$. Hence, we have $e \\in H \\cap K$, and so $H \\cap K$ satisfies the identity requirement for subgroups. \n\nFurthermore, $H \\cap K$ is a normal subgroup, because for all $a \\in G$, $a^{-1}(H \\cap K)a = a^{-1}Ha \\cap a^{-1}Ka = H \\cap K$. It's clear from the definition of intersection that $H$ and $K$ do not share a common subset larger than $H \\cap K$.\n \nFor $H, K \\in \\nsubg$, we have $H \\vee K = HK = \\{ hk \\mid h \\in H, k \\in K \\}$. \n\nFirst we will show that $HK$ is a group. For $hk, h'k' \\in HK$, since $kH = Hk$, there is some $h'' \\in H$ such that $kh' = h''k$. Hence, $hkh'k' = hh''kk' \\in HK$, and so $HK$ is closed under $G$'s group action. For $hk \\in HK$, we have $(hk)^{-1} = k^{-1}h^{-1} \\in k^{-1}H = Hk^{-1} \\subseteq HK$, and so $HK$ is closed under inversion. Since $e \\in H$ and $e \\in K$, we have $e = ee \\in HK$. Finally, $HK$ inherits its associativity from $G$.\n\nTo see that $HK$ is a normal subgroup of $G$, let $a \\in G$. Then $a^{-1}HKa = Ha^{-1}Ka = HKa^{-1}a = HK$.\n\nThere is no subgroup $F$ of $G$ smaller than $HK$ which contains both $H$ and $K$. If there were such a subgroup, there would exist some $h \\in H$ and some $k \\in K$ such that $hk \\not\\in F$. But $h \\in F$ and $k \\in F$, and so from $F$'s group closure we conclude $hk \\in F$, a contradiction.\n\n%%\n',
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