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Math 541 \\
Problem Set 6
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\item[3.1.1.]  $\phi(1) = 1 \in E$, so $1 \in \phi^{-1}(E)$, so $\phi^{-1}(E) \ne \varnothing$.  If $x, y \in \phi^{-1}(E)$ then $\phi(x), \phi(y) \in E$, so $\phi(xy^{-1}) = \phi(x)\phi(y)^{-1} \in E$ since $E$ is a subgroup, so $xy^{-1} \in \phi^{-1}(E)$.  Thus $\phi^{-1}(E)$ is a subgroup by the subgroup criterion.

Now suppose that $E$ is normal.  For all $g \in G$, 
\[ g\phi^{-1}(E)g \subseteq \phi^{-1}(\phi(g\phi^{-1}(E)g)) = \phi^{-1}(\phi(g)E\phi(g)^{-1}) = \phi^{-1}(E), \]
so $\phi^{-1}(E)$ is normal by Theorem 6, part (5).

The trivial subgroup $1 \le H$ is normal and $\ker \phi = \phi^{-1}(1)$, so $\ker \phi$ is normal.


\item[3.1.5.] The order of $gN$ is the smallest positive integer $\alpha$ such that $(gN)^\alpha$ is the identity $1N \in G/N$, but $g^\alpha N = 1N$ if and only if $g^\alpha \in N$, so the order of $gN$ is the smallest positive integer such that $g^\alpha \in N$.

For the example, take $G = \ZZ/6\ZZ$ and $H = \{0,3\}$; then the element 1 has order 6, but in $G/H = \ZZ/3\ZZ$, 1 has order 3.



\item[3.2.6.] Suppose that $Hg = kH$ for some $k \in G$.  Observe that $g \in Hg$ and $g \in gH$, so $gH$ intersects $kH = Hg$.  But two left cosets are either disjoint or equal, so $gH = kH = Hg$.  Multiplying on the right by $g^{-1}$, we have $gHg^{-1} = H$, so $g \in N_G(H)$.


\item[3.2.18.] Let $\phi: G \to G/N$ be the natural projection.  Then $\phi(H) \le G/N$, so $|\phi(H)|$ divides $|G/N| = |G:N|$.  But $\phi(H) \cong H/(H \cap N)$, so $|\phi(H)|$ divides $|H|$, so $|\phi(H)| = 1$, so $\phi(H) = 1$, so $H \subseteq \ker \phi = N$.


\item[3.3.3.] Let $\phi: G \to G/H$ be the natural projection, so as before $|\phi(K)|$ divides $|G:H| = p$.  Either $|\phi(K)| = 1$, so $K \subseteq \ker \phi = H$; or $|\phi(K)| = p$, so $\phi(K) = G/H$, so $G = HK$, and thus $G/H = HK/H \cong K/(K \cap H)$ by the diamond isomorphism theorem, so $|K : K \cap H| = |K/(K \cap H)| = |G/H| = |G:H| = p$.


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