Read: Section 7.1.
Turn in: 7.5(c), 7.7, 7.11, 7.15
- On 7.5(c), you want to prove that the set \(\phi(H)\) (defined in the problem) is a subgroup. You should use our usual method for doing this, and when doing so, you can make use of 7.5(a,b) even though you’re not being asked to prove them.
- Make sure to read 7.6; it’s a useful one.
-
For 7.7, if $|g| = m$, then you know that $g^m = e$ (and that $m$ is the smallest positive integer with this property); try applying $\phi$ to both sides. Corollary 4.25 should help.
-
For 7.11, there are two things to prove: (1) \(\ker(\phi)\) is a subgroup of \(G_1\) (Theorem 7.5(a,b) helps with this), and (2) \(\ker(\phi)\) is a normal subgroup of \(G_1\) (try Theorem 5.36 for this).
- Please prove 7.15 WITHOUT using 7.13 Showing that “$\phi$ is one-to-one $\implies$ \(\operatorname{ker}(\phi) = \{e_1\}\)” should follow quickly from the definition of the kernel. To prove “\(\operatorname{ker}(\phi) = \{e_1\}\) $\implies$ $\phi$ is one-to-one”, you should begin by assuming that $\phi(x) = \phi(y)$ (and you want to show $x=y$). To make use of your hypothesis that \(\operatorname{ker}(\phi) = \{e_1\}\), try manipulating the equation $\phi(x) = \phi(y)$ to get the identity on one side.
Extra practice: 7.5, 7.13
- To work with the condition “$a\in b\ker(\phi)$” in 7.13, it may help to remember what we learned about cosets before: $a\in b\ker(\phi) \iff a\ker(\phi) = b\ker(\phi) \iff b^{-1}a\in \ker(\phi) \iff a^{-1}b\in \ker(\phi).$ You probably won’t use all of those equivalent conditions, but some might prove useful. Now, when getting started, try manipulating the equation $\phi(a) = \phi(b)$ to get the identity on one side.