Let g be a group. suppose the number of elements in group G of order 5 is 28. determine the number of distinct subgroup of order 5.?

Question

Answer

Answer

To determine the number of distinct subgroup of order 5, we need to use the Sylow theorems.

Sylow Theorems:

For any prime p that divides the order of a finite group G, there exists a subgroup of G of order p^k for some k.

All subgroups of G of order p^k are conjugate to each other.

The number of subgroups of G of order p^k is congruent to 1 mod p and divides the order of G.

Using the Sylow theorems, we can determine the number of distinct subgroups of order 5 as follows:

Determine the number of Sylow 5-subgroups:

Since there are 28 elements of order 5, there must be at least one subgroup of order 5.

Let n be the number of Sylow 5-subgroups in G.

By the third Sylow theorem, n is congruent to 1 mod 5 and divides the order of G.

Therefore, n must be either 1 or 6 (since 6 is the only other number that is congruent to 1 mod 5 and divides 140, the order of G).

Determine if the number of Sylow 5-subgroups is 1 or 6:

If n = 1, then there is only one subgroup of order 5, and it is normal in G.

If n = 6, then there are six distinct subgroups of order 5, but they are not necessarily normal.

Therefore, the number of distinct subgroups of order 5 in G is either 1 or 6, depending on whether there is only one Sylow 5-subgroup or six Sylow 5-subgroups in G.

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