Quiz 2

NAME:

Part I:Answer the questions in the space provided. You do not need to show your work. Each problem is worth 5 points.

Quiz 2

1. Let $K=\{ \epsilon, (12)(34), (13)(24), (14)(23)\}$. How many distinct cosets of $K$ are there in $S_4$?

   
   

   The number of distinct cosets is the index $\vert S_4:K\vert$. By Lagrange's theorem, since $S_4$ is finite, this is $\vert S_4\vert/\vert K\vert=4!/4=6$.

   
   

2. Let $R$ be any commutative ring. Which one of the maps $\phi:R\to R$ defined below is a ring homomorphism? Circle one:
   • $\phi(r)=rk$ for some fixed element $k\in \mathbb{Z}$, $k\neq 0$, $k\neq 1$.
   • $\phi(r)=r+s$ for some fixed element $s\in R$, $s\neq 0$, $s\neq 1$.
   • $\phi(r)=re$ for some fixed idempotent $e\in R$, $e\neq 0$, $e\neq 1$.

   The completely correct answer to this question is NONE OF THE ABOVE. None of these maps preserves the multiplicative identity.

   However, the third map preserves addition since
   

   $$\phi(r+s)=(r+s)e=re+se=\phi(r)+\phi(s)$$
   
   
   and preserves multiplication since
   
   $$\phi(rs)=rse=rse^2=rese=\phi(r)\phi(s).$$
   
   
   So, the third map is a GENERALIZED ring homomorphism.

   I gave credit for either answer.

3. Show that every non-zero $z\in \mathbb{C}$ is a unit by finding the inverse for $z=a+bi$.

   If $a+bi\neq 0$ then $a^2+b^2\neq 0$ so $a/(a^2+b^2)$ and $b/(a^2+b^2)$ are real numbers. Then the multiplicative inverse of $a+bi$ is $({a\over a^2+b^2})- ({b\over a^2+b^2})i$. Here's another way of seeing that:

   
   

   $$(a+bi)(a-bi)({1\over a^2+b^2})=a^2+b^2/a^2+b^2=1$$
   
   

   The main point is that given a complex number, you just need to remember that multiplying by the complex conjugate gives you a real number. In almost EXACTLY the same way, you can show that the quatternions ${\mathbb{H}}$ form a field.

   Part II: Show all work. Each problem is worth 15 points. The best 3 of 4 problems below will count towards your grade.

4. Recall that $D_n$ is the group $\langle a, b\vert a^n=1, b^2=1, aba=b\rangle$. Show that $\langle a\rangle \vartriangleleft D_n$ for any $n\geq 2$.

   The following is a very short and simple proof:

   The subgroup $\langle a \rangle$ makes up exactly one half of $D_n$: elements in $D_n$ are either $a^k$ or else $a^kb$, and these are in one to one correspondence. Therefore, $\vert D_n:\langle a \rangle\vert=2$. By Theorem 4 in section 2.8 (this was also covered in class), that means that $\langle a \rangle$ is normal in $D_4$.

   Alternatively, you could show directly that $\langle a \rangle$ is normal in $D_4$. Since $D_4$ has generators $a$ and $b$, it is sufficient to show that
   

   $$a\langle a\rangle a^{-1}\subset \langle a\rangle$$
   
   
   (very easy) and also that
   
   $$b\langle a\rangle b^{-1}\subset \langle a \rangle$$
   
   
   (a little harder). This last is done by noting that if $ba^kb^{-1}\in b\langle a\rangle b^{-1}$ then using the relation $aba=b$, we obtain
   
   $$a^{-k} (a^k b a^k )b^{-1}=a^{-k} (b) b^{-1}= a^{-k}\in \langle a \rangle.$$
   
   

5. Let $G$ be a finite group and $H$ a normal subgroup of $G$. Prove that the order of the element $Hg$ in $G/H$ must divide the order of the element $g\in G$.

   One way to do this is by using Theorem 2 of Section 2.4. Let $\vert g\vert=n$. Then
   

   $$(Hg)^n=H(g^n)=H$$
   
   
   which is the identity element of $G/H$. So... since $(Hg)^n=e_{G/H}$, we have the order of the element $Hg$ divides $n$ by part (a) of Theorem 2.

   You did not need to quote the theorem number to get full credit.

6. For any group $G$, prove that $G/Z(G)$ is isomorphic to $Inn(G)$. (Recall: $Z(G)$ is the center of $G$, and $Inn(G)$ is the group of automorphisms of the form $\sigma_a(g)=aga^{-1}$.)

   Whenever you are faced with proving that two things are isomorphic, it's always a good idea to try to use the Isomorphism theorem.

   Define a map $f:G\to Inn(G)$ by $f(a)=\sigma_a$ for $a\in G$. Then $f$ is a homomorphism:

   
   

   $$f(ab)=\sigma_{ab}=\sigma_a\circ \sigma_b=f(a)f(b)$$
   
   

   since $\sigma_{ab}(g)=(ab)g(ab)^{-1}=a(bgb^{-1})a^{-1}=a\sigma_b(g)a^{-1} =\sigma_a(\sigma_b(g))$ for all $g\in G$.

   In fact, $f$ is onto since given any $\sigma_a\in Inn(G)$, we have $\sigma_a=f(a)$.

   Now, $c\in Ker(f)$ if and only if $f(c)=1_G$, the identity map of $G$. That is, $c\in Ker(f)$ iff $\sigma_c=1_G$. For all $g\in G$, this implies that
   

   $$\sigma_c(g)=cgc^{-1}=g=1_G(g)$$
   
   
   and so $cg=gc$ for all $g\in G$. That is, $c\in Ker(f)$ iff $c\in Z(G)$.

   By the isormophism theorem, $G/Z(G)\cong Inn(G)$.

7. Let $\phi:R_1\to R_2$ be a ring homomorphism. Show that $ker(\phi)$ is an ideal in $R_1$.

   Recall that $ker(\phi)=\{r\in R_1 \vert \phi(r)=0\}$. This was one of the things that I did not anticipate that people would be confused about.

   To show that $ker(\phi)$ is an ideal we must show that $ker(\phi)$ is an additive subgroup and $ker(\phi)$ is left and right closed under multiplication. Additive subgroup:

   • $0\in ker(\phi)$ since $\phi(0)=0$.

   • S'pose that $a,b\in ker(\phi)$. Then $a+b\in ker(\phi)$ since $\phi(a+b)=\phi(a)+\phi(b)=0+0=0$.

   • S'pose that $a\in ker(\phi)$. Then $-a\in ker(\phi)$ since $\phi(-a)=-\phi(a)=-0=0$.

   Finally, if $r\in R$ and $a\in ker(\phi)$, then $ar\in ker(\phi)$ since
   

   $$\phi(ar)=\phi(a)\phi(r)=0\phi(r)=0$$
   
   
   and similarly $ra\in ker(\phi)$ since
   
   $$\phi(ra)=\phi(r)\phi(a)=\phi(r)0=0.$$