Homework 25

Read: Read Section 7.1.

Turn in: 7.7, 7.9, 7.11, 7.15, 7.16(a,b), 7.20

• For 7.7, if $|g| = m$, then you know that $g^m = e$ (and that $m$ is the smallest positive integer with this property); if you’re stuck, 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 $G_1$, and (2) $a\ker(\phi) = \ker(\phi)a$ for all $a\in G_1$.

• 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.

• On 7.16, the idea is that if you know that $\phi(i)= h$ and $\phi(j) = v$, then, for example, you can figure out $\phi(k) = \phi(ij) = \phi(i)\phi(j)=hv$.

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.