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1To prove that (G, ⊗ ) is commutative group, we need to show 1)G is closed with respect to our operation, 2) our operation is associative, 3)there is an identity in G 4)every element in G has inverse in G, and 5) any two elements commute. 1) 1Clearly the table shows that when we combine any two elements in G, we alway

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1To prove that (G,
⊗
) is commutative group, we need to show 1)G is closed withrespect to our operation, 2) our operation is associative, 3)there is an identity in G 4)everyelement in G has inverse in G, and 5) any two elements commute.1) 1Clearly the table shows that when we combine any two elements in G, we always getan element in G.2) 1Since we know our operation is composition, therefore we have associative property. 3) From the table there is an element d of G that is the identity. Because every elementcombined with d gives back that element. So, G has an identity.4) Looking at our table again we can see our identity d is positioned in each element on thediagonal. This means each element is its own inverse. So, each element of G has an inversein G.5) 1Clearly we can see that our table is symmetric with respect to the main diagonal. Thisshows us that any two elements in H commute.Therefore, we can conclude that we have a commutative group. Now let’s look for any nontrivial subgroup of G. From the table we can find twosubgroups of the group G. One of the subgroups is P= {a, b, c, d}. To prove that (P,
⊗
) issubgroup of the group G, we need to show (P,
⊗
) is a group and
P G
∅≠ ⊆
. Then(P,
⊗
) is a subgroup of (G,
⊗
) if and only if 1) P is closed with respective operation;2)
e P
∈
, where
e
is the identity element of G; 3).
a P
∈
implies
1
a P
−
∈
, where
1
a
−
is theinverse of a in G.
1)
Clearly the table shows that when we combine any two elements in P, we alwaysget an element in P.
2)
From the table the element d is the identity in P.
3)
Looking at our table again, we can see each element of P is its own inverse.Therefore, we can conclude (P,
⊗
) is subgroup of the group G.Another subgroup of (G,
⊗
) is the cyclic subgroup that is generated by the element (v, d).((v, d))= {(v, d),(
0
360
r
, d)}. This subgroup has order two, meaning that 2 is the smallest positive integer exponent on (v, d) that is needed to reach the identity.The next goal is finding normal subgroups. To prove our subgroup is normal wecan show 1) A subgroup P is normal in the group G if and only if aP = Pa for every
a G
∈
or 2) The subgroup P is a subgroup of the commutative group G.1) We know P is a subgroup of the group of G and P= {a, b, c, d}. First we need pick up anelement from G, say a. Then we can calculate the left coset (a x P) and right coset(P x a).
a x P = a x {a, b, c, d} = {(a x a), (a x b), (a x c), (a x d)}= {d, c, b, a}P x a = {a, b, c, d} x a = {(a x a), (b x a), (c x a), (d x a)}= {d, c, b, a} Now we can see, (a x P) = (P x a) for
a G
∈
.
Repeating this process for every element in G,we will see that our subgroup P is normal with two distinct coset.2) Theorem 2.45(p134) and Corollary can support the fact that omit P is a normalsubgroup of the G. Theorem 2.45 states that if P is a subgroup of the group G such that
P centG
⊆
, then P is a normal ( commutative) subgroup of G and the corollary statesevery subgroup of a commutative group is normal. Since our group of G is commutativegroup then we can conclude that our subgroup is normal by the corollary. To prove thecorollary, we can suppose that (G,
⊗
) is a commutative group, and
P G
∅≠ ⊆
. Since G iscommutative, the center of G which is centG= {
a G
∈
/ a x b= b x a for all
b G
∈
} , isequal to the set G. As we know P is a subgroup of G then P is also a subset of centG.Therefore by thereom2.45 P is normal subgroup of G.The last part of our discussion is to look for a nontrivial factor group for our group.Let’s use the group G and the normal subgroup P of G.Then G / P= {a x P /
a G
∈
}. On
G / P define, operation#by (aP)(bP) = ( ab)( P). Now we need to show 1) G/P is closure; 2) [(aP)#(bP)]#(cP) = (aP)#[ (bP)#(cP)];3) eP= P
∈
G/P ; 4) (aP)#(
1
a
−
P)= eP, (
1
a
−
P)#(aP)=eP.The table for the operation G/H .
#[a x P][b x P][a x P][d x P][c x P][b x P][c x P][d x P]
1) Clearly from the table, G/P is closed with respective #.2) Suppose aP, bP, cP
∈
G/P.[(aP) # (bP)] # (cP) = (cP) # (cP)= (d P)= (aP) # (aP)= (aP) # [(bP) # (cP)]Repeating this process for every element in G/P, we will verify associative property.3) From the table, there is an identity dP in G/P(a P) # (d P) = (a d) P= a P(d P) # (a P) = (d a) P = a P4) From the table we can see that each element is its own inverse. Now we can conclude our group is factor group.

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