1. ## subgroup question

Let H be a subgroup of G. Need to prove that Z(G) intersect H is contained in Z(H) (Z= center of the group, i.e the subset of elements in G that commute with every element of G). From this result I then need to verify that Z(G) intersect H is a normal subgroup of H. I am unsure how to do the first part of this so any help would be great, for the second part is it that Z(G) intersect H is normal in H because anything that is commutative in all of G will also be commutative in H?

2. If an element of H commutes with everything in G, then surely it commutes with everything in H (because H is a subgroup of G). It's trivial to show that the center is normal, since $aba^{-1}=baa^{-1}=b$ when a and b commute.

3. would I have to show that Z(G) intersect H is a subgroup of H by a subgroup test and then use a normaility test to show that Z(G) intersect H is normal in Z(H). Or could I use double containment to show that Z(G) intersect H equals Z(H), and then get that Z(G) intersect H is a nromal subgroup of H?

4. "would I have to show that Z(G) intersect H is a subgroup of H by a subgroup test and then use a normaility test to show that Z(G) intersect H is normal in Z(H)."

Here is an easy result that will make life easy:
Intersection of subgroups is a subgroup in both the subgroups. I wrote a proof but latex is annoying me i'll posts in a bit.

be careful tiny boss it would make no sense to prove Z(G) intersect H was normal in H if Z(G) intersect H wasn't a group itself.

With that result you can just show that if a is an element of both Z(G) and H then it is in the group Z(H), and use the normality test which says if H is normal in G then for any element x in G xHx^-1 is a subset of H. Since any element from Z(G) and H is in Z(H), it would mean for any element x from H and for any element a from Z(G) intersect H, xax^-1 = xx^-1a since a is commutative with all elements from H, so that equals a which is in Z(G) intersect H. so Z(G) intersect H is normal in H

5. a. wts Z(G) intersect H is contained in Z(H)
So Z(G)=a is an element of G|ax=xa for all x that is an element of G. So Z(G) intersect H = a is an element of H|ax=xa for all x that is an element of G. Since an element a of Z(G) intersect H is in H and commutes with all of G, then it obviously commutes with all of H since H is a subgroup of G. Thus Z(G) intersect H is contained in H.

b. LTS that Z(G) intersect H is a subgroup of H.
Obviously Z(G) intersect H is contained within H as shown above.
Since the identity e is always in Z(G) and in H (since H is a subgroup of G), then Z(G) intersect H is nonempty. Let a,b be elements of Z(G) intersect H. Then ab=ba by defintion of Z(G) intersect H. Let a(b^-1) be an element of Z(G) intersect H.
Then a(b^-1)x=(b^-1)ax (since ab=ba implies (b^-1) = (a^-1)(b^-1)a.
Thus, by the 1-step subgroup test, Z(G) intersect H is a subgroup of H.
Since Z(G) intersect H = a is an element of H|ax=xa for all x that is an element of G,
then Z(G) intersect H, then the element a obviously commutes with everything in H if it commutes with everything in G. Thus Z(G) intersect H is an abelain subgroup of H and is normal.

6. The proof is quite trivial but here it is anyway

Intersection of two subgroups is always a subgroup in both the groups and thus also in G.

Assume H and M are subgroups of a group G
Assume a, b $\in H \cap M$ By def of intersection, b $\in H \: and \: b \in M$ Since inverses in groups are unique, $b^{-1} \in H \: and \: b^{-1} \in M \: so b^-1 \in H \cap M \: and \: by \: closure \: ab^{-1} \in H \: and \: ab^{-1} \in M$ We are done. Since $H \cap M$ is a subset of both H and M, and additionally is a group, it is a subgroup in H, M and G.