# Math Help - Modern Algebra, Groups, Integral domains, fields...

1. ## Modern Algebra, Groups, Integral domains, fields...

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 $\phi : I50 \to I50$ is a group isomorphism $\phi(7) = 13$.
Determine $\phi(x)$

I was trying to find generator of I50 all I found is that $3^{15}= 7$ and that $3^{17}=13$ but $<3>$ 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 $g \to g^2$ is an automorphism of G.

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

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

My answer: Given $H={g^5|g \in G}$ is a subgroup of G we must show that $xHx^{-1} \subset H, \forall x \in G$
Let $x \in G$ and $y^5 \in H$ for some $z \in G$.
Since $y^5 \in H$ then by definition $y \in H$
Let $z = xyx^{-1}$, with $xyx^{-1} \in G$ according to the definition of a conjugate.
Then let $z^5=(xyx^{-1})^5 = (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1})$
$= xy(x^{-1}x)y(x^{-1}x)y(x^{-1}x)y(x^{-1}x)yx^{-1}=xy^5x^{-1}$
By definition $z^5 \in H$, thus $xy^5x^{-1} \in H$.
Therefore H is a normal subgroup.

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

My answer: Let $F4 = { \begin{bmatrix}x & y\\x & (x+y)\end{bmatrix} , x,y \in F2}$
$F4 = { \begin{bmatrix}0&0\\0&0\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}1&1\\1&0\end{bmatrix}, \begin{bmatrix}0&1\\1&1\end{bmatrix}}$

F4 is a field of 4 elements because it is a commutative ring (close under multiplication and addition), it has a unity $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ 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 $a,b \in R$. Suppose $a^3 = b^3$ and $a^7=b^7$. Prove that $a= b$

My answer: $a^7=b^7$
$a.a^6=b.b^6$
$a.(a^3)^2= b.(b^3)^2$
$a=b$ since $a^3=b^3$.

2. Originally Posted by ynn6871
2. Suppose that G is a finite Abelian Group and G has no element of order 2. Show that the mapping $g \to g^2$ is an automorphism of G.
Define $\phi : G\to G$ by $\phi (g) = g^2$.
Since $G$ is finite it is sufficient to prove that $\phi$ is injective for $\phi$ to be a bijection.
If $\phi(g_1) = \phi(g_2) \implies g_1^2 = g_2^2 \implies (g_1g_2^{-1})^2 \implies g_1g_2^{-1} = e\implies g_1=g_2$
And $\phi(g_1g_2) = (g_1g_2)^2 = g_1^2g_2^2 = \phi (g_1) \phi (g_2)$.

3. Le G be a group of order $p^n$ where p is prime. Prove that the center G cannot have order $p^{(n-1)}$.
Say $Z(G)$ has order $p^{n-1}$.
Pick $a\in G - Z(G)$ and construct $C(a) = \{x\in G | xa = ax \}$.
Now prove that $C(a)$ is a subgroup of $G$.
Notice that $Z(G)$ is properly contained in $C(a)$.
This forces $|C(a)| = p^n$ because $Z(G) \subset C(a) \subseteq G$ by Lagrange's theorem.
But then it must mean that $a$ commutes with all of $G$.
Therefore, $a\in Z(G)$. A contradiction!

3. Hi.

For 2), verfying that $\Phi_{2}:g\mapsto g^{2}$ is a group morphism isn't hard because of commutativity. What is $Ker(\Phi_{2})$? The set of elements in $G$ whose square is the identity, i.e. the identity and all elements of order $2$. So $\Phi_{2}$ is injective and, since $G$ is finite, bijective.

For 3), let $Z(G)$ be the center of $G$. With the hypothesis, as $Z(G)$ is normal in $G$, $G/Z(G)$ order is $p$, so is a cyclic group. Hence $G$ is abelian, and $Z(G)=G$. Contradiction.

4. For number 2. can you tell me what are the implication of
G has no element of order 2
??? Did you use it to prove it is a bijection???

5. Originally Posted by ynn6871
4. Let G be a group. If $H={g^5|g \in G}$ is a subgroup of G, prove that it is a normal subgroup of G.

My answer: Given $H={g^5|g \in G}$ is a subgroup of G we must show that $xHx^{-1} \subset H, \forall x \in G$
Let $x \in G$ and $y^5 \in H$ for some $z \in G$.
Since $y^5 \in H$ then by definition $y \in H$
Let $z = xyx^{-1}$, with $xyx^{-1} \in G$ according to the definition of a conjugate.
Then let $z^5=(xyx^{-1})^5 = (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1})$
$= xy(x^{-1}x)y(x^{-1}x)y(x^{-1}x)y(x^{-1}x)yx^{-1}=xy^5x^{-1}$
By definition $z^5 \in H$, thus $xy^5x^{-1} \in H$.
Therefore H is a normal subgroup.
I do not like how you wrote up your solutions. Let $h\in H$ and $g\in G$ we wish to show $ghg^{-1} \in H$. Notice that $h=x^5$ for some $x\in G$. Therefore, $ghg^{-1} = gx^5g^{-1} = (gxg^{-1})^5 \in H$.

5.Show that there is an integral domain with exactly 4 elements.
Many ways to answer (all of which are isomorphic), one can be found here.

??? Did you use it to prove it is a bijection???
Yes. Because if $g_1^2 = g_2^2 \implies g_1 = g_2$ since there is no element of order 2.

6. I am sorry but I am still having a hard time understanding the no elements of order 2.
Does it mean that since there exists no elements of order 2 then $g^2=g$?? I just don't understand the concept behind, I am sorry I am having a hard time with it all!

Also do you have any suggestion as to how to approach problem number 1 ??
1. Suppose that is a group isomorphism .
Determine
Clic-clac: when you said For 3),
let be the center of . With the hypothesis, as is normal in , order is , so is a cyclic group. Hence is abelian, and . Contradiction.
, can I just assume that is normal in G or do I have to prove it somehow, the rest of your proof makes total sense to me (I think) it's just that one part.
Let me rephrase the solution really quick:

We assume that Z(G) has order $p^{n-1}$ and that it is normal in G, then G/Z(G) has order p, since $p / p^{n-1}$ is equal to p. Then since p is prime then G is abelian, and Z(G)=G which is a contradiction. But what happens in the case that Z(G) is not normal, is Z(G) ever not normal???

Thank you very much for your help guys, I am getting really confused in this class!!!

7. Originally Posted by ynn6871
I am sorry but I am still having a hard time understanding the no elements of order 2.
Does it mean that since there exists no elements of order 2 then $g^2=g$?? I just don't understand the concept behind, I am sorry I am having a hard time with it all!
Define the function $\phi : G\to G$ by $\phi (x) = x^2$. Since $G$ has no element of order two it means that if $x^2 = e$ then $x=e$. Therefore, to show that $\phi$ is one-to-one we need to show that if $\phi (a) = \phi( b )$ then $a=b$. Now if $a^2 = b^2$ then $a^2b^{-2} = e$ and so $(ab^{-1})^2 = e$ because the group is abelian. But that means $ab^{-1} =e$ and finally $a=b$.

Let me rephrase the solution really quick:

We assume that Z(G) has order $p^{n-1}$ and that it is normal in G, then G/Z(G) has order p, since $p / p^{n-1}$ is equal to p. Then since p is prime then G is abelian, and Z(G)=G which is a contradiction. But what happens in the case that Z(G) is not normal, is Z(G) ever not normal???
Try to prove the following theorem: If $H$ is normal in $K$ so that $K/H$ is cyclic and $H$ is abelian then $K$ has to be abelian. You will realize this is a special case when $K=G$ and $H = Z(G)$. But how do we show that $Z(G)$ is normal. That is easy! Let $g\in G$ then $gZ(G) = Z(G)g$ because $Z(G)$ contains all elements which commute with everyone. We see that the left cosets of $Z(G)$ are equal to the right cosets of $Z(G)$. That is one of the equivalent conditions for a subgroup to be normal. Thus, $Z(G)$ is normal.

8. Any suggestion on how to go about problem 1???

9. What is $I50$?

10. Integers modulo 50

11. Ok! Then, in $I50$, we get:

$7^{2}=-1\ ,$ so $7^{4}=1$.

$\phi(1)=\phi(7^{3}\times 7)=7^{3}\times \phi(7)=-7\times 13=-41=9$

So $\phi:I50\rightarrow I5O:n\mapsto 9n$

12. Originally Posted by clic-clac
$I50$, we get:

$7^{2}=-1\ ,$ so $7^{4}=1$.

$\phi(1)=\phi(7^{3}\times 7)=7^{3}\times \phi(7)=-7\times 13=-41=9$

So $\phi:I50\rightarrow I5O:n\mapsto 9n$

Can you explain your thought process a little bit??? Why did you pick powers of 7??? And why did you start with $\phi(1)$????
Will this function be a bijection???? is everything mapped?

13. Well a group endomorphism $f$ of $I50$ will be determined by the range of $1$, which is a generator, and would be like : $n\mapsto f(1)n$ (that's easy to show with morphism properties).

Here, we can see that $7^{2}\equiv -1\ (50)$. So $7^{4}\equiv 1\ (50)$.

Then I just calculate $\phi(1)$ (modulo 50 of course).

As you may see, with that result, we have $\phi(7)=7\phi(1)=7\times 9=63 \equiv 13\ (50)$, that sounds not too bad.

Finally, $pgcd(9,50)=1$, so $9$ is a generator of $I50$, hence your morphism is bijective.

Ps. Since $7$ and $13$ are also relative primes with $50$, $\phi(7)=\phi(13)$ means the range of a generator is a generator, with morphism properties, that implies $\phi$ is an isomophism!

14. $
\phi(1)=\phi(7^{3}\times 7)=7^{3}\times \phi(7)
$
Ok I understand where you are coming from a little better, the only other thing I don't understand is how can you just pull out the $7^3$ outside of $\phi()$ ??? shouldn't it be something like $
\phi(1)=\phi(7^{3}\times 7)=\phi(7)\times\phi(7)\times\phi(7)\times\phi(7)= 11
$
which gives you $\phi(n)=11n$but then I don't get the desired $\phi(7) = 13$ but $\phi(7) = 27$ which is not what I want.

15. Oh that's a tempting error

The group is $(I50,+)$, so $\phi(a+b)=\phi(a)+\phi(b)$ $\\$(not $\phi(ab)=\phi(a)\phi(b)$ !)

Therefore $\phi(n\times7)=\phi(7+7+...+7)=\phi(7)+\phi(7)+... +\phi(7)=n\phi(7)$

In our case, $n=7^{3}$