# Suppose $G$ is a group with $|G|=35$. Prove that if $H$ is a subgroup of $G$ with order 7, then $H$ is a normal subgroup of $G.$

(1) Suppose $G$ is a group with $|G|=35$. Prove that if $H$ is a subgroup of $G$ with order $7$, then $H$ is a normal subgroup of $G.$

(2) Suppose that G is a group with $|G|=35.$ Prove that if $G$ is abelian, then $G$ must be a cyclic group.

Ok, so here is what I have so far. We have to suppose to that $G$ is a group and that $H$ is a subgroup of finite index. Let $n=[G:H]$. We know to be true that there exists a homomorphism $\varphi:G \to S_n$ s.t. $Ker(\varphi) \subseteq H.$

Suppose that $G$ is a group of order $35$ and that $H$ is subgroup of order $7$. Then we have, $n=[G:H]=|G|/|H|=35/7=5$ By the Cayley Homomorphism theorem, there exists $\varphi :G \to S_5$ s.t. $Ker(\varphi) \subseteq H$. Let $K=Ker(\varphi)$. Thus, $K$ is a subgroup of $H$ and by Lagrange's theorem, $|K|$ divides $|H|=7$, therefore $|K|=1$ or $7$.

If $|K|=1$, then $\varphi$ would be injective and therefore $G$ would be isomorphic to $\varphi(G),$ a subgroup of $S_5$ which means that $S_5$ would have a subgroup of order $35$ and by Lagrange, $S_5 \mid 35$ but $|S_5|=120$ and $120 \not \mid 35$. Therefore, $|K| \neq 1$.

That proves $|K|=7$ and since $K \subseteq H$ and $|H|=7,$ it follows that $H=K$ and since $K$ is the kernel of a homomorphism, $K$ must be a normal subgroup of $G$. Hence $H$ is a normal subgroup of $G$.

I am hoping that this is correct. I am not sure what else I need or how to go on to prove (2).

• Do you have the fundamental theorem of finitely generated abelian groups at your disposal for 2? You don't need it, but it renders it pretty much trivial. – Alex Wertheim Mar 1 '15 at 21:46
• We can always let $G$ act on left cosets of $H$ by $g\cdot xH = gxH$. This yields a homomorphism $G \to S_5$ and the kernel must be contained in $H$ since only $H$ fixes the coset $H$. – David Wheeler Mar 1 '15 at 22:10
• @DavidWheeler, are you referring to acts on a group?? This concept was only just introduced to me, I am not so familiar with it – cele Mar 1 '15 at 22:23
• Yes! Typically these are used in the proofs of various Cayley-esque theorems, although not necessarily referenced as such (for example, Herstein glosses over group actions, and just uses the homomorphisms). A group action is just a certain kind of group homomorphism, $\phi$, from a group $G$ to a symmetric group. The image $\phi(g) = \phi_g: x \mapsto \phi_g(x)$ is often written $x \mapsto g\cdot x$, so that a group action can be thought of as a way to "multiply" a group element, and a set element (much like the "scalar product" is a way to multiply a vector by a scalar). – David Wheeler Mar 1 '15 at 22:30
• For (2): your group is essentially $\mathbb{Z}_{35}$, which is isomorphic (by the Chinese remainder theorem, or by hand) to $\mathbb{Z}_5 \times \mathbb{Z}_7$, of which $(1,1)$ is a generator. – Alex M. Mar 20 '15 at 16:23

Using Sylow's theorems we can arrive at the result. For starters, in any group of order $35$ there is a subgroup of order $7$ in $G$ by Sylow's first theorem. By Sylow's third theorem we have that the number of Sylow subgroups of order $7$,say $x$ for simplicity, is congruent to $1$ modulo $7$. So $x=1+7k$ for some $k$. But $x$ also divides $[G:H]$ by Sylow's third theorem. Thus $x|5$ leaving that there is only $1$ subgroup of order $7$. Finally, we conclude that $H$ is normal by Sylow's second theorem, which states that all Sylow $p$ subgroups are conjugate to one another. Since there is only one subgroup of order $7$, we have $gHg^{-1}=H$ for every $g\in G$.
Suppose $|G|=35$ and that $G$ is an abelian group. By Cauchy's theorem for finite groups we have that $G$ contains subgroups of order $5$ and $7$. Say $o(x)=5$ and $o(y)=7$. Since $G$ is abelian, we have $o(xy)=\operatorname{lcm}(x,y)=35$. As $G$ contains an element of order $35$, $xy$, we have that $G$ is cyclic, generated by $xy$.