# Abstract Algebra

• Oct 15th 2008, 08:59 PM
GreenandGold
Abstract Algebra
I have no clue how to attack these question one. All help or hints is appreciated.

G always denotes a group

1. Let G be finite and G not equal to {e}. Show that G has an element of prime order.

2. Prove that isomorphic groups have isomorphic automorphism groups.

3. Let a, b be in G . If |a| and |b| are relatively prime (i.e. gcd (|a|, |b|)=1), then <a> intersection <b> ={e}. Prove the last statement.

Thanks!
• Oct 15th 2008, 09:30 PM
ThePerfectHacker
Quote:

Originally Posted by GreenandGold
1. Let G be finite and G not equal to {e}. Show that G has an element of prime order.

Let $\displaystyle a\in G - \{ e \}$ and construct $\displaystyle \left< a \right>$. Let $\displaystyle |\left< a \right> | = n>1$. The integer $\displaystyle n$ has a prime divisor $\displaystyle p$. The element $\displaystyle a^{n/p}$ has order $\displaystyle p$.

Quote:

2. Prove that isomorphic groups have isomorphic automorphism groups.
Let $\displaystyle \phi : G_1 \to G_2$ be a group isomorphism.
Define $\displaystyle \hat \phi : \text{Aut}(G_1)\to \text{Aut}(G_2)$ by $\displaystyle \hat \phi (\theta)(a) = \phi(\theta(a))$.
Show this is an isomorphism.

Quote:

3. Let a, b be in G . If |a| and |b| are relatively prime (i.e. gcd (|a|, |b|)=1), then <a> intersection <b> ={e}. Prove the last statement.
Notice that $\displaystyle \left< a\right> \cap \left< b \right>$ is a subgroup of $\displaystyle \left< a \right>$ and $\displaystyle \left< b\right>$. By Lagrange's theorem it means $\displaystyle | \left< a\right> \cap \left< b\right>|$ divides $\displaystyle |\left< a\right>| = |a|$ and $\displaystyle \left< b\right>| = |b|$. Therefore, $\displaystyle |\left< a\right> \cap \left< b \right>| = 1 \implies \left< a \right> \cap \left< b \right> = \{ e \}$.
• Oct 16th 2008, 01:30 AM
GreenandGold
Quote:

Originally Posted by ThePerfectHacker
Let $\displaystyle a\in G - \{ e \}$ and construct $\displaystyle \left< a \right>$. Let $\displaystyle |\left< a \right> | = n>1$. The integer $\displaystyle n$ has a prime divisor $\displaystyle p$. The element $\displaystyle a^{n/p}$ has order $\displaystyle p$.

Let $\displaystyle \phi : G_1 \to G_2$ be a group isomorphism.
Define $\displaystyle \hat \phi : \text{Aut}(G_1)\to \text{Aut}(G_2)$ by $\displaystyle \hat \phi (\theta)(a) = \phi(\theta(a))$.
Show this is an isomorphism.

Notice that $\displaystyle \left< a\right> \cap \left< b \right>$ is a subgroup of $\displaystyle \left< a \right>$ and $\displaystyle \left< b\right>$. By Lagrange's theorem it means $\displaystyle | \left< a\right> \cap \left< b\right>|$ divides $\displaystyle |\left< a\right>| = |a|$ and $\displaystyle \left< b\right>| = |b|$. Therefore, $\displaystyle |\left< a\right> \cap \left< b \right>| = 1 \implies \left< a \right> \cap \left< b \right> = \{ e \}$.

So need to show that $\displaystyle \hat \phi (\theta)(a) = \phi(\theta(a))$ is a structure preserving function?
• Oct 16th 2008, 07:26 AM
ThePerfectHacker
Quote:

Originally Posted by GreenandGold
So need to show that $\displaystyle \hat \phi (\theta)(a) = \phi(\theta(a))$ is a structure preserving function?

Yes