Let G=U(32) and K= {1,15}.
Prove that G/K is isomorphic to Z8.
I know that |G/K|= |U(32)|/|K| = 16/2=8.
G/K= {K,3K,5K,7K,17K,19K,21K,23K}.
While trying to see that the properties of isomorphic groups work on G/K and Z8, I tried to find the number of elements of order 8 in G/K. I found that 3K, 5K, 19K and 21K have order 8 in G/K like 1,3,5 and 7 in Z8.
I know that this can't prove isomorphism but really I can't approach the solution. Maybe the Cayley Table may help but not in PROVING I think.
Frankly, I'm not strong enough at isomorphism or normal groups as my professor illustrated them quickly in like 2 lectures, and I have a final exam.
Answer
\newcommand{\Int}{\mathbf{Z}}Let n be a positive integer. A cyclic group (G, \cdot) of order n is isomorphic to the "standard" cyclic group (\Int_{n}, +): If a \in G is a generator, then every element of G may be written in the form a^{r} for a unique integer r with 0 \leq r < n, and the group law is
a^{r} \cdot a^{s} = a^{r + s} = a^{r + s \pmod{n}}
by the law of exponents and the fact that a^{n} = e in (G, \cdot).
In fancy language, the mapping \phi:\Int_{n} \to G defined by \phi(r) = a^{r} induces an isomorphism of groups (\Int_{n}, +) \to (G, \cdot) because
\phi(r + s) = a^{r + s} = a^{r} \cdot a^{s} = \phi(r) \cdot \phi(s)
for all integers r and s.
In your example,
\begin{alignat*}{2} (3K)^{1} &= \{ 3, 13\} &&= 3K, \\ (3K)^{2} &= \{ 9, 7\} &&= 7K, \\ (3K)^{3} &= \{27, 21\} &&= 21K, \\ (3K)^{4} &= \{17, 31\} &&= 17K, \\ (3K)^{5} &= \{19, 29\} &&= 19K, \\ (3K)^{6} &= \{23, 25\} &&= 23K, \\ (3K)^{7} &= \{ 5, 11\} &&= 5K, \\ (3K)^{8} &= \{15, 1\} &&= K. \end{alignat*}
This shows the mapping \phi:\Int_{8} \to U(32)/K defined by \phi(r) = (3K)^{r} induces an isomorphism. (I'm being picky by saying "induces" rather than "is": The mapping acts on sets, while isomorphism is a property of groups—sets equipped with group operations.)
The other generators you found induce distinct isomorphisms, four in all.
No comments:
Post a Comment