Let $G$ be a finite group which does not have two subgroups of the same order (cardinality of the group). Assuming that I need to show that
- any subgroup of the group $G$ is its normal divisor (normal subgroup), done
- if $G$ is nontrivial, then it's isomorphic with the product of its Sylow subgroups (hint Induction in regard of amount of Sylow subgroups),
- $G$ is cyclic (hint Use $2.$).
Starting with $1.$ I wanted to assume that there exists a subgroup $N$ of $G$ s. t. $gN\neq Ng$ for every $g \in G$. Namely, it is a subgroup which is not normal. Using $3.$ we obtain
$$gN\neq Ng,$$ $$g^nN\neq Ng^n,$$ $$eN\neq Ne,$$ $$N\neq N.$$
which makes a contradition. Anyway, I have a problem with $2.$ and $3.$. Any hints would be helpful.
I know that the third problem was somehow solved here as one user mentioned, but I am not convinced that this is the same situation. Anyway, the second problem with isomorphism still exists.
Thanks.
As for the second claim, since the Sylow subgroups are normal, and conjugate, there is only one of each. Then say for starters there are only two, $P$ and $Q$. Their intersection is trivial by Lagrange, since their orders are relatively prime. And $G=PQ$, by counting, and by Lagrange again. Now proceed with the induction.
For $3$, it follows from the fact that $(m,n)=1\implies\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{mn}$, a well-known fact, following from the Chinese remainder theorem.
Next see https://www.researchgate.net/publication/279181950_A_Characterization_of_the_Cyclic_Groups_by_Subgroup_Indices. This result shows that the Sylow subgroups, under our assumption, will be cyclic. Thus we can apply the previous result.