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!!

Hint: see here http://math.stackexchange.com/questions/255441/how-can-i-prove-that-every-group-of-n-255-elements-is-commutative/255474#255474. Try to adapt. — Alex Youcis, Jun 06 '13 at 21:59

score 2 · Answer 1 · answered Jun 06 '13 at 22:10

2

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

answered Jun 06 '13 at 22:10

Jack Schmidt

55,589

thanks a lot, but I cannot see the Journal, without registering for the site – AnonymouseCat Jun 07 '13 at 07:56
The deeper knowledge may remain hidden until you visit the library. The hints I gave do not require these secrets, only the textbook Sylow and isomorphism theorems. – Jack Schmidt Jun 07 '13 at 12:57
$+1^{\infty}$ I always enjoy reading your neat answers which are based on old documents and books. This is the right way of teaching abstract concepts. When I had a course about infinite abelian groups, you answered some of my questions. Thanks. Just saying thank you. :-) – Mikasa Jun 09 '13 at 06:03

score 1 · Answer 2 · answered Jun 06 '13 at 23:02

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.

Why is $Aut(N)\cong C_6 \times C_12$? and what does $C_i$ mean? — AnonymouseCat, Jun 07 '13 at 07:58
$;C_n=$ the (multiplicative, usually) cyclic group of order $;n;$. In general, the automorphism group of a direct product of cyclic groups is the direct product of the automorphisms groups. — DonAntonio, Jun 07 '13 at 09:11

Finding a subgroup in the Center with order 91

2 Answers2