answer:First, we note that there is no non-abelian group of order 10, since the only non-abelian group of order 2p, where p is an odd prime, is the dihedral group Dp. Therefore, we only need to consider whether G is isomorphic to C10 or D5. By the Sylow theorems, we know that the number of 2-Sylow subgroups in G divides 5 and is congruent to 1 modulo 2. Thus, there can be either 1 or 5 such subgroups. Similarly, the number of 5-Sylow subgroups in G divides 2 and is congruent to 1 modulo 5, so there must be exactly one 5-Sylow subgroup. Let P be the unique 5-Sylow subgroup of G. Since P is the only 5-Sylow subgroup, it is normal in G. If there is only one 2-Sylow subgroup Q, then Q is also normal in G. In this case, G has a normal subgroup of order 2 and a normal subgroup of order 5. Since the orders of P and Q are coprime, their intersection is trivial, and thus G is isomorphic to the direct product of P and Q, which is a cyclic group of order 10, i.e., G is isomorphic to C10. If there are 5 distinct 2-Sylow subgroups, none of them are normal in G. In this case, G cannot be a cyclic group, since a cyclic group of order 10 has a unique subgroup of order 2. Therefore, G must be isomorphic to the dihedral group D5. In conclusion, G is either isomorphic to the cyclic group C10 or the dihedral group D5. To determine which one, we can examine the number of 2-Sylow subgroups in G. If there is only one, G is isomorphic to C10, and if there are five, G is isomorphic to D5.

answer:We will analyze each of the given groups and compare their properties with the given group G. 1. Cyclic group C10: A cyclic group of order 10 has a generator element of order 10. However, in the given group G, every element except the identity element has order 2. Therefore, G cannot be isomorphic to the cyclic group C10. 2. Dihedral group D5: The dihedral group D5 has order 10, but it has elements of order 2 and 5. In the given group G, every element except the identity element has order 2. Therefore, G cannot be isomorphic to the dihedral group D5. 3. Non-abelian group of order 10: Since every element of G except the identity element has order 2, G must be an abelian group. This is because for any two elements a and b in G, we have (ab)^2 = abab = aabb = ee = e, where e is the identity element. Thus, ab = ba, which means G is abelian. Therefore, G cannot be isomorphic to a non-abelian group of order 10. Since G is not isomorphic to any of the given groups (C10, D5, or the non-abelian group of order 10), we conclude that G must be a different group. One such group that satisfies the given conditions is the Klein four-group V extended by a cyclic group of order 2, which is an abelian group of order 10 where every element except the identity element has order 2.

answer:The non-abelian group of order 10 is actually the dihedral group D5. There is no other non-abelian group of order 10. So, we only need to compare the given multiplication table with the tables for C10 and D5. The cyclic group C10 has the following composition table: | * | e | a | b | c | d | f | g | h | i | j | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | e | e | a | b | c | d | f | g | h | i | j | | a | a | b | c | d | e | f | g | h | i | j | | b | b | c | d | e | f | g | h | i | j | a | | c | c | d | e | f | g | h | i | j | a | b | | d | d | e | f | g | h | i | j | a | b | c | | f | f | g | h | i | j | a | b | c | d | e | | g | g | h | i | j | a | b | c | d | e | f | | h | h | i | j | a | b | c | d | e | f | g | | i | i | j | a | b | c | d | e | f | g | h | | j | j | a | b | c | d | e | f | g | h | i | Comparing the given table with the C10 table, we can see that they are not the same. Therefore, the group G is not isomorphic to C10. Now, let's compare the given table with the dihedral group D5. The dihedral group D5 has the following composition table: | * | e | a | b | c | d | f | g | h | i | j | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | e | e | a | b | c | d | f | g | h | i | j | | a | a | b | c | d | e | g | h | i | j | f | | b | b | c | d | e | f | h | i | j | a | g | | c | c | d | e | f | g | i | j | a | b | h | | d | d | e | f | g | h | j | a | b | c | i | | f | f | j | i | h | g | e | d | c | b | a | | g | g | f | j | i | h | d | e | b | c | a | | h | h | g | f | j | i | c | b | e | d | a | | i | i | h | g | f | j | b | c | d | e | a | | j | j | i | h | g | f | a | b | c | d | e | Comparing the given table with the D5 table, we can see that they are not the same either. Therefore, the group G is not isomorphic to D5. Since the group G is not isomorphic to either C10 or D5, and there is no other non-abelian group of order 10, we can conclude that the given group G is not isomorphic to any of these groups.

answer:To prove this, we will use the Sylow Theorems. Recall that the order of a group G is the number of elements in the group. The Sylow Theorems give us information about the possible subgroups of a group based on the order of the group. Since the order of G is 10, we can write it as a product of its prime factors: 10 = 2 * 5. Now, we will apply the Sylow Theorems to find the possible Sylow subgroups of G. 1. Sylow 2-subgroups: By the Sylow Theorems, the number of Sylow 2-subgroups (subgroups of order 2) must divide the order of the group and be congruent to 1 modulo 2. In this case, the possibilities are 1 and 5. If there is only one Sylow 2-subgroup, it must be normal in G. 2. Sylow 5-subgroups: Similarly, the number of Sylow 5-subgroups (subgroups of order 5) must divide the order of the group and be congruent to 1 modulo 5. In this case, the only possibility is 1. So, there is only one Sylow 5-subgroup, and it must be normal in G. Now, let P be the unique Sylow 5-subgroup of G, and let Q be a Sylow 2-subgroup of G. Since P is normal in G, we have that PQ is a subgroup of G. Moreover, since the orders of P and Q are relatively prime, we have that the order of PQ is |P||Q| = 5 * 2 = 10. Thus, PQ = G. If there is only one Sylow 2-subgroup, then G is the internal direct product of P and Q, and G is isomorphic to the cyclic group C10. This is because P and Q are both cyclic (since they have prime order), and their orders are relatively prime. If there are five Sylow 2-subgroups, then there must exist an element x in G that is not in any of the Sylow 2-subgroups. This element x cannot commute with every element of G, because if it did, it would be in the center of G, and G would be abelian. But in this case, G would be isomorphic to the cyclic group C10, which contradicts the assumption that there are five Sylow 2-subgroups. Therefore, if there are five Sylow 2-subgroups, there exists an element in G that does not commute with every element of G. In this case, G must be isomorphic to either the dihedral group D5 or the non-abelian group of order 10, since these are the only non-abelian groups of order 10. In conclusion, either G is isomorphic to the cyclic group C10, or there exists an element of G that does not commute with every element of G, and hence G must be isomorphic to either the dihedral group D5 or the non-abelian group of order 10.