I have an assgnment due on Thursday in Mordern Algebra and I must say that I am very lost I got (I think) 3 problem out of the 6 and I was wondering if someone could help me out and check the work I have done already.

Ok here it is, any suggestion for 1, 2 and 3????

1. Suppose that $\displaystyle \phi : I50 \to I50 $ is a group isomorphism $\displaystyle \phi(7) = 13 $.

Determine $\displaystyle \phi(x)$

I was trying to find generator of I50 all I found is that $\displaystyle 3^{15}= 7$ and that $\displaystyle 3^{17}=13$ but $\displaystyle <3>$ doesn’t map all of I50 so I am not sure what to do.

2. Suppose that G is a finite Abelian Group and G has no element of order 2. Show that the mapping $\displaystyle g \to g^2$ is an automorphism of G.

3. Le G be a group of order $\displaystyle p^n$ where p is prime. Prove that the center G cannot have order $\displaystyle p^{(n-1)}$.

4. Let G be a group. If $\displaystyle H={g^5|g \in G}$ is a subgroup of G, prove that it is a normal subgroup of G.

My answer: Given $\displaystyle H={g^5|g \in G}$ is a subgroup of G we must show that $\displaystyle xHx^{-1} \subset H, \forall x \in G$

Let $\displaystyle x \in G$ and$\displaystyle y^5 \in H$ for some $\displaystyle z \in G$.

Since $\displaystyle y^5 \in H$ then by definition $\displaystyle y \in H$

Let $\displaystyle z = xyx^{-1}$, with $\displaystyle xyx^{-1} \in G $ according to the definition of a conjugate.

Then let $\displaystyle z^5=(xyx^{-1})^5 = (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1})$

$\displaystyle = xy(x^{-1}x)y(x^{-1}x)y(x^{-1}x)y(x^{-1}x)yx^{-1}=xy^5x^{-1}$

By definition $\displaystyle z^5 \in H$, thus $\displaystyle xy^5x^{-1} \in H$.

Therefore H is a normal subgroup.

5.Show that there is an integral domain with exactly 4 elements.

My answer: Let $\displaystyle F4 = { \begin{bmatrix}x & y\\x & (x+y)\end{bmatrix} , x,y \in F2}$

$\displaystyle F4 = { \begin{bmatrix}0&0\\0&0\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}1&1\\1&0\end{bmatrix}, \begin{bmatrix}0&1\\1&1\end{bmatrix}} $

F4 is a field of 4 elements because it is a commutative ring (close under multiplication and addition), it has a unity $\displaystyle \begin{bmatrix}1&0\\0&1\end{bmatrix}$ and each non zero elements has a multiplicative inverse. Since F4 is a field then it is an integral domain with 4 elements.

6.Let R be an integral domain and $\displaystyle a,b \in R$. Suppose $\displaystyle a^3 = b^3$ and $\displaystyle a^7=b^7$. Prove that $\displaystyle a= b$

My answer: $\displaystyle a^7=b^7$

$\displaystyle a.a^6=b.b^6$

$\displaystyle a.(a^3)^2= b.(b^3)^2$

$\displaystyle a=b$ since $\displaystyle a^3=b^3$.