Finding a subgroup in the Center with order 91 Question:
Let G be a group of order $455=5\cdot 7\cdot 13$. 


*

*Show that exists a normal subgroup $ H<G: |H|=91$ and $H\subseteq Z(G)$.

*Show that G is an Abelian and cyclic group.


Solution:
So I showed that exists a normal subgroup by using Sylow's 3rd theorem to show that exists only one subgroup of order 7 $H_7$ (which is normal) and only one subgroup of order 13, $H_{13}$ (which is normal as well according to Sylow's 3rd). Then, I showed that $H_7 \cap H_{13}={e}$, such that $H_7\cdot H_{13}$ is a normal subgroup of order $7\cdot 13=91$. 
From here, I didn't really know how to show that $H_7\cdot H_{13}$ is in the Center and that G is Abelian and cyclic. 
Thanks a lot for the help!!
 A: For a 4th solution:
How many Sylow 5 subgroups does $G/H_7$ have?
How many Sylow 5 subgroups does $G/H_{13}$ have?
Every subgroup of a quotient $G/H_i$ is of the form $K_i/H_i$ for some subgroup $K_i$ of $G$. How big is $K_7 \cap K_{13}$?
Is it normal?
This exercise is constructed in a silly way. Neither 7 nor 13 is equivalent to 1 mod 5, so of course the Sylow 5-subgroups are normal, by Hall (1928). However most students are not taught Hall's results, and so that they have to reprove them in smaller situations like this.


*

*Hall, P.
“A note on soluble groups.”
Journal of the London Mathematical Society 3, (1928) 98-105.
JFM 54.0145.01
DOI:10.1112/jlms/s1-3.2.98
A: You also could try the following: put $\,P_r\,$ for a Sylow $\,r$-subgroup, then:
A group $\,G\,$ of order $\;455=5\cdot 7\cdot 13\;$ has one unique Sylow $\,13$-subgroup, from which it follows that 
$$P_{13}\lhd G\implies N:=P_{13}P_7\le G\;$$
and since $\,[G:N]=5=\,$ the minimal prime dividing $\,|G|\,$ , we get that in fact $\,N\lhd G\,$ , so that our group is an extension of  $\,N\,$ by a (any) Sylow $\,5$-subgroup $\,P_5\,$. But
$$\text{Aut}(N)\cong C_6\times C_{12}\implies|\text{Aut}(N)|=72$$
and since $5\nmid 72\,$ , the only possible homomorphism $\,P_5\to\text{Aut}(N)\,$ is the trivial one, from where the corresponding semidirect product is in fact direct:
$$N\rtimes P_5=N\times P_5$$
and since $\,N\,$ is abelian (and in fact cyclic: why?) and since also $\,P_5\;$ is,  we're done.
