Math 120A HW 9 Solutions

1. Some examples:
   1. $aH = bH$
   2. $a^{-1}bH = H$
   3. $a^{-1}bH \subseteq H$
   4. $a^{-1}b \in H$.
   To see that condition a implies condition b, left-multiply both sides by $a^{-1}$. Clearly condition b implies condition c. To see that condition c implies condition d, note that $a^{-1}b = a^{-1}be \in a^{-1}bH$ where $e$ denotes the identity element of $G$.

   To see that condition d implies condition a, assume that condition d holds. Then $b^{-1}a \in H$ also, because $H$ is closed under inverses. Now consider an arbitrary element $ah$ of $aH$, where $h \in H$. Then $ah = (b(b^{-1}a))h = b((b^{-1}a)h) \in bH$. Therefore $aH \subseteq bH$. For the converse, consider an arbitrary element $bh$ of $bH$, where $h \in H$. we have $bh = (a(a^{-1}b))h = a((a^{-1}b)h) \in aH$. Therefore $bH \subseteq aH$.

   (I don't remember why I put the hypothesis of normality in the question.You could use it to get equivalent statements like $aH = Hb$, or $aHb^{-1} = H$, but I don't think these are especially interesting. Perhaps I meant to ask about conditions equivalent to the normality of $H$; you should know several of those also.)

2. What we need to show is that if $a,b\in H$ and $aH = bH$, then $\phi(a) = \phi(b)$. So assume that $a,b \in H$ and $aH = bH$. Then $a^{-1}b \in H = \ker(\phi)$, so $\phi(a)^{-1}\phi(b) = \phi(a^{-1}b) = e'$ where $e'$ denotes the identity element of $G'$. Multipying both sides on the left by $\phi(a)$, we get $\phi(b) = \phi(a)e' = \phi(a)$ as desired.
3. See the back of the textbook.
4. See the back of the textbook.
5. See the back of the textbook.