GATE Overflow Test Series | Discrete Mathematics | Test 3 | Question: 10

Question

Let $(G,\ast)$ be a group of order $P$, where $P$ is a prime number. The number of subgroups of $G$ is ________

• A. $0$
• B. $P^{2}$
• C. $P^{2}-2$
• D. $2$

Answer 1

As per Lagrange's theorem, order of a subgroup must divide the order of the group. Since, the order of the group $p$ is prime, there is only $2$ possible orders for the subgroups -- $p$ and $1$ -- both giving the trivial subgroups.

So, the correct answer is $(D)$.

If the question had asked for "nontrivial" subgroups, answer would have been $0.$

Answer 2

This can also be resolved by sylow’s 1^st theorem.  It states that if the order of group G is divisible by any prime number p^k where “k” is any power, then all the subgroups of G will be of the order of p^k.

Hence   |G| /p^k

according to question,  |G| /p^k =    p/p = 1

Hence all the subgroups H of G are of the order p only.   Now there can be only 1 subgroup where |H|= |G| and another subgroup with |H|=NULL.  Thus there can be 2 such subgroups. Option D is correct