About the first Sylow theorem The first Sylow theorem states the existence of $p$-subgroups. I was wondering if there can be other subgroups of the same order and different structure.
Let's take a big group which has a $2^2$ factor in its order. The theorem guarantees it will have at least one $2$-subgroup of order $4$.
Let's assume, thanks to the third Sylow theorem, we find $N$ of these subgroups, which turn to be Klein $4$ groups. Now the second Sylow theorem is telling us they will be all Klein $4$ groups since they are conjugated.
The question is: can this group contain other subgroups of order $4$ with a different structure, like $C_4$ for instance?
Thanks in advance.
 A: The wording of your question is a little murky. If $2^2$ is the largest power of $2$ dividing the order of your group, then "No" - if one subgroup of order 4 is a Klein 4-group, then all must be since subgroups of order 4 here are Sylow 2-subgroups and thus all isomorphic. On the other hand, consider $G=C_2 \times C_2 \times C_4$ (the direct product of two cyclic groups of order 2 and a cyclic group of order 4). In this case $G$ is its own (unique) Sylow 2-subgroup. Also, $G$ contains a subgroup with Klein 4-group structure and a subgroup with cyclic (order 4) structure.
What can be said in general? If finite group $G$ contains a subgroup $H$ of prime power order (say $p^k$), then every Sylow $p$-subgroup of $G$ must itself have a subgroup isomorphic to $H$.
So if you have a finite group with subgroups isomorphic to both $C_2 \times C_2$ and $C_4$, then that group must have Sylow 2-subgroups large enough to accommodate both of these structures. For example, the dihedral group of order 8 is such a group.
A: Yes and no.
Say, let $G$ be a finite group and let $p$ be a prime factor of $n=|G|$ and let $k$ be the highest power of $p$ dividing $n$. Then for all $i=1,2,\ldots, k$ there will be subgroups of $G$ order $p^i$, including for $i=k$. Those don't need to be isomorphic. Say, in $G=S_4$ (the group of permutations of the set $\{1, 2, 3, 4\}$) you can have a subgroup of order $4$ generated by the cycle $(1 2 3 4)$ (isomorphic to $C_4$) and another subgroup of order $4$ generated by two cycles $(1 2)$ and $(3 4)$ (which is isomorphic to Klein group).
However, for the special case $i=k$: any two subgroups of order $p^k$ are conjugated - hence isomorphic. To continue with the previous example: as $|S_4|=4!=24=2^3\cdot 3$, any two subgroups of order $2^3=8$ in it are necessarily isomorphic to each other.
