Math 3124 Friday, November 14
Third Sample Test Solutions
1. (a) h(123)i={(1),(123),(132)}and
((1 2)(3 4))(123)((1 2)(3 4))−1= (214)/∈ h(123)i.
Therefore h(123)iis not normal in H.
(b) If |H|=6, then [H:h(1 2 3)i] = 2, consequently h(1 2 3)iH(remember, sub-
groups of index 2 are always normal) contradicting the above. Therefore |H| 6=6.
(c) Note that o((1 2 3)) = 3 and o((1 2)(3 4)) = 2. Therefore by Lagrange’s theorem,
|H|is a multiple of 3 and 2, consequently |H|is a multiple of 6. Also |H| 6=6 by
the above, and |H||24, again by Lagrange’s theorem. The only possibilities left
are |H|=12 and |H|=24. But |H|=24 tells us that H=S4, which is certainly
normal in S4, while |H|=12 tells us that [S4:H] = 2 and again HS4.
Remark Actually one can show that |H|=12.
2. The general element of G/His of the form Hg where g∈G. Then (Hg)5=H(g5) =
H. Therefore the order of Hg divides 5, so must be 1 or 5. But if the order is 1, then
Hg must be the identity element of G/H. It follows that every nonidentity element of
G/Hhas order 5.
3. The right cosets are a:=H([0],[0]) ={([0],[0]), ([0],[2]), ([2],[0]), ([2],[2]) }
b:=H([0],[1]) ={([0],[1]), ([0],[3]), ([2],[1]), ([2],[3]) }
c:=H([1],[0]) ={([1],[0]), ([1],[2]), ([3],[0]), ([3],[2]) }
d:=H([1],[1]) ={([1],[1]), ([1],[3]), ([3],[1]), ([3],[3]) }
The Cayley table is ⊕a b c d
a a b c d
b b a d c
c c d a b
d d c b a
4. (a) Let A,B∈G. Using det(AB) = det(A)det(B), we obtain θ(AB) = det(AB)2=
det(AB)det(AB) = det(A)det(B)det(A)det(B) = det(A)2det(B)2=θ(A)θ(B), so
θis a homomorphism. If x∈P, then we may write x=y2for some y∈P. We
now have
x=θy0
0 1,
and the above matrix is in G, hence θis onto.
(b) A∈kerθ⇐⇒ det(A2) = 1⇐⇒ det(A)2=1⇐⇒ det A=±1⇐⇒ A∈N. There-
fore kerθ=N.
(c) This now follows from the fundamental homomorphism theorem.
