# Thread: Abstract Algebra Help!!!

1. ## Abstract Algebra Help!!!

Hey im preparing for my second exam and I need major help with these problems. I would greatly appreciate any help, hints or answers, that could help me understand how to do these problems. So that I can do well thanks!!!

G denotes a group

1. Let |G|=p*g*r with different primes p, g, r, and let H,K<=G with |H|=p*g, |K|=g*r. Show:|H (intersection) K|=g.

2. Let H be the normal subgroup G, a in G and |H|=10. The element aH of the group G/H has order 3; what are the possibilities for |a|? Explain.

3. Let G be a group and G' be the subgroup generated by S={x^-1*y^-1*x*y| x,y in G}. Then:

a. Prove that G' is normal in G.
b. Prove that G/G' is abelian.
c. If G/N is abelian, prove that G'<=N.
d. Prove that if H<=G and G'<=H then H is a subgroup of G.

4. Suppose that G is an abelian group of order 120 and that G has exactly 3 elements of order 2. Determine the isomorphism class of G. Explain.

2. Originally Posted by GreenandGold
1. Let |G|=p*g*r with different primes p, g, r, and let H,K<=G with |H|=p*g, |K|=g*r. Show:|H (intersection) K|=g.
If $H,K$ are subgroups of a finite group $G$ then $|HK|\cdot |H\cap K| = |H||K|$.

Now since $H\cap K$ is a subgroup of $H$ and $K$ it means its order divides $|H| = pg$ and $gr$. Therefore, $|H\cap K| = 1\text{ or }g$. We cannot have $|H\cap K| = 1$ for that would imply $pgr=|G| \geq |HK| = |H||K| = pg^2r$. A contradiction.

Therefore, the only possible case is that $|H\cap K| = g$.

2. Let H be the normal subgroup G, a in G and |H|=10. The element aH of the group G/H has order 3; what are the possibilities for |a|? Explain.
If $|aH| = 3$ in the group $G/H$ it means $(aH)^3 = H$ and therefore $a^3 \in H$. But $|H| = 10$. Therefore, $(a^3)^{10} = a^{30} = 1$. Since $a\not = 1$ it means the possible values for $|a|$ are $2,3,5,6,10,15,30$.

3. Originally Posted by GreenandGold
3. Let G be a group and G' be the subgroup generated by S={x^-1*y^-1*x*y| x,y in G}. Then:

a. Prove that G' is normal in G.
b. Prove that G/G' is abelian.
c. If G/N is abelian, prove that G'<=N.
d. Prove that if H<=G and G'<=H then H is a subgroup of G.
For $x,y\in G$ define $[x,y] = xyx^{-1}y^{-1}$ then show that $g[x,y]g^{-1} = [gxg^{-1},gyg^{-1}]$.
Therefore, we see that $G'$ is a normal subgroup.

Since this group is normal we can form the factor group $G/G'$. Let $aG',bG' \in G/G'$. To show $(aG')(bG') = (bG')(aG')$ it is necessary and sufficient to prove that $(ab)(ba)^{-1} \in G' \implies aba^{-1}b^{-1} \in G'$. This is true by definition of $G'$.

If $G/N$ is abelian it means $(aN)(bN) = (bN)(aN)$ and so $aba^{-1}b^{-1} \in N$. This is true for all $a,b\in G$. Therefore, $G'\subseteq N$.

4. Originally Posted by GreenandGold
4. Suppose that G is an abelian group of order 120 and that G has exactly 3 elements of order 2. Determine the isomorphism class of G. Explain.
Let $P$ be the Sylow $2$-subgroup of order $8$.

Consider, $G/P$. This group has order $15$ and therefore $G/P \simeq \mathbb{Z}_{15}$ because that is the only group up to isomorphism.

Since $|P| = 8$ and it is abelian implies that $P$ must be isomorphic to one of the following groups: $\mathbb{Z}_8, \mathbb{Z}_2^3, \mathbb{Z}_4\times \mathbb{Z}_2$.
Elements of order $2$ must be contained in $P$. Therefore, by hypothesis, we are told that $P$ has exactly $3$ elements of order $2$.
The first two groups (on the list) do not satisfy this requirement.
However, $\mathbb{Z}_4\times \mathbb{Z}_2$ has that requirement with elements $(0,1),(2,1),(2,0)$.

Therefore, $G/P \simeq G/(\mathbb{Z}_4 \times \mathbb{Z}_2) \simeq \mathbb{Z}_3 \times \mathbb{Z}_5$.

Now we "cross multiply" and we get, $G\simeq \mathbb{Z}_2\times \mathbb{Z}_3 \times \mathbb{Z}_4 \times \mathbb{Z}_5$.