Question

Name: 1. Prove that if N is a subgroup of index 2 in a group G, then N is normal in G 2. Let N < SI consists of all those permutations ơ such that o(4)-4. Is N nonnal in sa? 3. Let G be a finite group and H a subgroup of G of order . If H is the only subgroup of G of order n, then is normal in G 4. Let G be a group, and kt C-ra e Glar-ra for all reG} be the center of G. Show that C is normal subgroup of G 5. Consider the subgroups < 6 > and < 30 > of Z·Show that < 6 >/< 30 > 6. Consider the group G = Z16 with 11 = < 9 > and K =< 18 >. Zs. (a) List the cosets in G/H, shoning the elements in each coset (b) List the cosets in G/K, showing the elements in each coset (e) List the cosets in H/K, showing the elements in each coset (d) List the cosets in (G/K)/(H/K), showing the elements in each coset e Show that (G/H) (G/K)/(H/K). [Define an isomorphism between G/H and (G/k)(H/K) 7. Let ф : G-, G , where G is any group, be given by Og)-9-1, for g e G. Determine whether the map φ is a hornomorphisan. 8. Let ф : G- G, be a honomorphisin with kernel 11 and let a E G. Iet S-{r e G e(1)- H@ and HaS s). φ(a)]. Prove S = Ha. (Hint Show that S 9. List all elements of Zs x Z6 and the elements of the cyelie group < (0,1) >. Then compute the factor group(Z x 26)/< (0,1) > 10. Let φ: Zis·→ Z12 be the limomorphismi wiereo(1)-10. (a) Find K-ker() (b) List the cosets in Zjs/K showing the elements in each coset (c) Find ф1z,a]. 0 495

Answer Question 5 .

Answer 1

Let $G=<6>$ and $H=<30>$ . Since is an abelian group, is a normal subgroup of . Therefore is a group ( with elements of the form , where $g \in G$ ).

The elements of are:

$0+H=H=\{0,\pm 30,\pm 60,\pm 90,...\}$,

$6+H=\{...,-84,-54,-24,6,36,66,96,...\}$,

$12+H=\{...,-78,-48,-18,12,42,72,102,...\}$,

$18+H=\{...,-72,-42,-12,18,48,78,108,...\}$ and

.

4H-...,-66,-36, -6, 24,54, 84, 114,...

See that is the zero element in .

We form the mapping $\phi : \mathbb{Z}_5\rightarrow G/H$ by $\phi(\overline{k})=6k+H$ for $0\leqslant k\leqslant 4, k \in \mathbb{Z}$ .

So $\overline{0} \in \mathbb{Z}_5$ is mapped to the zero element in and $\phi (m+n)=6(m+n)+H=(6m+H)+(6n+H)=\phi(m)+\phi(n)$ from the rules of coset addition.   So $\phi$ is a homomorphism.

The map $\psi:G/H\rightarrow \mathbb{Z}_5$ given by $\psi(6k+H)=\overline{k}$ for $0\leqslant k\leqslant 4, k \in \mathbb{Z}$ is the inverse of $\phi$ and this shows that $\phi$ is bijective.

Hence $\phi$ is an isomorphism and so $<6>/<30>\cong \mathbb{Z}_5$ .

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