Let G be a group of order 17. The total number of non- isomorphic subgroups of G isa)1b)2c)3d)17Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Explanation:

To determine the number of non-isomorphic subgroups of a group G, we need to consider the possible orders of subgroups of G.

1. Lagrange's Theorem:
Lagrange's theorem states that the order of any subgroup of a group divides the order of the group itself. In this case, the order of the group G is 17.

2. Possible Orders of Subgroups:
The possible orders of subgroups of G are factors of 17, which are 1 and 17. Since every group has at least two subgroups, the trivial subgroup {e} and the group itself G, we can conclude that there are at least two subgroups of G.

3. Non-Isomorphic Subgroups:
To determine if the two subgroups are non-isomorphic, we need to examine their structures. Since the order of G is prime (17), G is a cyclic group generated by any of its elements. Let's say g is a generator of G.

3.1. Trivial Subgroup:
The trivial subgroup {e} is isomorphic to itself. It is a subgroup of every group.

3.2. Group Itself:
The group G is also isomorphic to itself. It is a subgroup of every group.

3.3. Subgroups of Order 17:
Since G is cyclic, it only has one subgroup of order 17, which is G itself.

3.4. Subgroups of Order 1:
Since G is cyclic, it has exactly one subgroup of order 1, which is the trivial subgroup {e}.

Therefore, the total number of non-isomorphic subgroups of G is 2 (trivial subgroup and the group itself), which corresponds to option B.

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