Use Langrange's Theorem to see if you can conclude that a proper subgroup H of G is cyclic

Question

Let $(G,•)$ be a group of order n. Let $H$ be a proper subgroup of $G$ where $H$ does not equal $G$. Use Langrange's theorem to see if you can conclude that $H$ must be cyclic.

$n=12$ Divisors: $1, 2, 3, 4, 6$

$n=25 $ Divisors: $1, 5$

$n=35$ Divisors: $1,5,7$

$n=29$ Divisors: $1$

So I think if a subgroup $H$ has an order (number of elements) equal to $G$ then $H=G$ so I omitted those divisors. Now off the bat i realize $n=29$ would mean we only have one proper subgroup which would be the identity and would thus be cyclic.

For the others, I dont quite get where to follow

Answer 1

Recall that a group of prime order is unique and must be cyclic. Hence, for $n=25$, proper subgroups can be of order $1$ and $5$; for $1$ the subgroup is the identity (as you said) and is hence cyclic and a subgroup of order $5$ has prime order so it must be cyclic as well. Similarly, for $n=35$, the proper subgroups are the trivial one and subgroups of order $5, 7$, both of which are prime and hence those subgroups are cyclic. You already stated the solution for $n=29$. For $n=12$, can you try to come up with a counter example? (Hint: consider $G=S_3\times C_2$).

Answer 2

Hint: Once you consider the possible orders of the subgroups, look at the divisors of the subgroups. In particular, which ones are necessarily cyclic?