Finding a subgroup in the Center with order 91

Question

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!!

Answer 1

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

Answer 2

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.