**1) If N is a subgroup of index 2 in a group G, then N is normal in G.
Proof-** Let be the set of distinct left cosets of N in G. Let be the set of distinct right cosets of N in G. N itself is always both a right and left coset of a subgroup N of G. Also, cosets partition the group so we can write:

Thus, every left coset of N in G is also a right coset of N in G so that N is normal in G.

**2) If is a family of normal subgroups of a group G, then is a normal subgroup of G.
Proof-** Let and This says that . Since each is normal we have that . Thus so that is a normal subgroup of G.

**3) Let N be a subgroup of a group G. N is normal in G iff congruence modulo N is a congruence relation.
Proof-** We know that congruence modulo N is an equivalence relation on G by previous exercise so we only need to show if . Assuming the hypothesis we have that and . This tells us that and . Then we have and for some . Then note:

. We have that because N is normal in G. Thus as well by closure of N. Therefore we have:

so that

Let . Since N is a subgroup of G, we can write so that . Also . Thus we can use the congruence relation property to say that:

. This tells us that which gives that N is normal in G.

**7) Let G be a finite group and H a subgroup of G with order n. If H is the only subgroup of G of order n, then H is normal in G.
Proof-** We know from previous exercise that and so . The uniqueness of H implies that which by definition gives H as normal in G.

**9)(a) If G is a group then the center of G is a normal subgroup of G.
Proof-** Note that the center of G is the set . This set is already a subgroup of G. Let . Then so the center is normal in G

**12) If H is a normal subgroup of a group G such that H and G/H are finitely generated, then so is G.
Proof-** Let and . Claim that . Note first that by closure of the group G. Let . Then so that we can write , where each for some and . Since H is normal in G, we have G/H is a group under multiplication so we can change . By definition of cosets this gives . Thus , where each for some and . It then follows that

**13) (a) Let H, K be normal in G. Show that is normal in G.
Proof-** Let . By definition of the join of H and K we can write where , and . Then:

. Now if a particular or we have that or since H and K are both normal in G. This each for any i which gives that . Thus the join of H and K is normal in G.

**14) If and then and .
Proof-** Let . Then because and . Thus .

Now we define by . This function is clearly well defined and surjective. Note that:

so that f is a homomorphism. Thus f is an epimorphism. Lastly, we have because whenever which can only happen when and . So by the first isomorphism theorem we have the conclusion:

.

**15) Let and . If and then
Proof-** Let be the canonical epimorphism and consider . The restriction of a group homomorphism to a group is still a group homomorphism. Let . Then and . So then so that the restriction is injective. Next, suppose that . Then so that . But since K and N are both normal in G we can write any for . Thus g can be written . Then so that the map restricted to K is surjective. Thus the map is an isomorphism so we are done.

**19) If , is finite, , is finite, and , then
Proof-** Let . Since H is finite so is . Consider . Note that . By Lagrange we also have that . Thus . Also, because the order of hN must divide . Therefore, is a common divisor of , but we know the gcd of these two numbers is 1 so that |hN| = 1, which tells us that hN is the identity element in G/N, i.e. so that . Thus H is a subgroup of N.