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How can I show that every group of order $35$ is abelian ? I know that if such a group is abelian the it's isomorphic to $\mathbb Z_{35}$ or $\mathbb Z_7\times \mathbb Z_5$. But, how can I show that any group of order 35 is either isomorphic to $\mathbb Z_{35}$ or to $\mathbb Z_7\times \mathbb Z_5$ ?

MSE
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    Have you covered Sylow theorems? Another remark: Can you show that $f:\Bbb{Z}_{35}\to\Bbb{Z}_5\times\Bbb{Z}_7, \overline{n}\mapsto (\overline{n},\overline{n})$ is an isomorphism? – Jyrki Lahtonen Jun 23 '16 at 21:02
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    Do you know any theorems about groups of order $pq$, where $p$ and $q$ are prime? – Alex Wertheim Jun 23 '16 at 21:02
  • I do, but unfortunetly, not for this exercise (since it's an exercise I had at the beginning of the semester). But what I know is that there is a element of order 5 and an other one of order 7. – MSE Jun 23 '16 at 21:04
  • I suppose you mean an element of order 5. – Bernard Jun 23 '16 at 21:06

3 Answers3

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SyLow implies that you have a subgroup of order $5$ $L$ and a subgroup on order $7$, $M$. The number $n_5$ of Sylow 5-groups divides 7 and is equal to 1 mod 5 so is 1 so $L$ is normal. Let $x$ be an element of order $7$, the conjugation by $x$ induces an automorphism of $L$ of order $5$, so is trivial since the order of $x$ is 7, thus if $y$ generates $L$, $x$ commutes with $y$ and the group is the product $Z_5\times Z_7$.

Tsemo Aristide
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  • What do you mean by "automorphism of order 5" ? And are you sure it's really an automorphism ? (because you can have that $gxg^{-1}=hxh^{-1}$ for $g\neq h$, no ? – MSE Jun 23 '16 at 21:19
  • It means that $f^5=Id_M$. Let $f([1])=[m]$, $f([x])=xf([1])=[mx]$, this implies that $f^5(x)=[m^5x]=[mx]$ since $m^5=m$ mod 1 by little fermat. – Tsemo Aristide Jun 23 '16 at 21:25
  • It is an automorphism since $g(xy)g^{-1}=gxg^{-1}gyg^{-1}$ – Tsemo Aristide Jun 23 '16 at 21:28
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    You can also see this without discussing automorphisms. Since $|L|$ and $|M|$ are coprime, $L \cap M = 1$, so $|LM| = |L||M|/|L \cap M| = |L||M| = 35 = |G|$, hence $G = LM$, so $G$ is a semidirect product of $L$ and $M$. We have already seen that $n_5 = 1$, and by the same reasoning, $n_7 = 1$, hence $L$ and $M$ are both normal, so the semidirect product is in fact direct. –  Jun 23 '16 at 21:29
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edit: I apologize, but during writing this argument I did not notice that Tsemo Aristide had anticipated me and had given a sketch of what seems to be exactly the same argument(with some minor differences, since I did not use Sylow's theorem). I am looking forward for comments should I delete my answer or leave it as other people may find almost complete argument I have written usefull.

Suppose that there are two distinct subgroups $H_1$, $H_2$ inside $G$ of order $7$. Then $H_1\cap H_2=\{1\}$, otherwise we would have $H_1=H_2$. Using the fact that $H_1\cap H_2=\{1\}$ we can derive that $H_1H_2$ has cardinality $7^2=49$. This is a contradiction. Thus there is at most one subgroup of $G$ of order $7$.

By Cauchy's theorem there exists a subgroup $H\subseteq G$ of order $7$. By what we have said above such subgroup is unique. Obviously such subgroup is normal(even characteristic).

Again by Cauchy's theorem there is a subgroup $K\subseteq G$ of order $5$. Since $H$ is normal we have homomorphism $\phi:K\rightarrow \mathrm{Aut}(H)$ given by: $$\phi(k)(h)=k^{-1}hk$$ Next note that $K$ is of order $5$ and: $$\mathrm{Aut}(H)\cong \mathrm{Aut}(\mathrm{Z}_7)=\mathrm{Z}_6$$ Thus there are no homomorphisms from $K$ to $\mathrm{Aut}(H)$ except for the trivial one. We deduce that $\phi$ is trivial. This means that: $$k^{-1}hk=h$$ for every $k\in K$ and $h\in H$. Therefore, elements of $H$ and $K$ commute with each other.

Since $H\cap K=\{1\}$, we derive that $G=HK$. Hence $G$ is abelian.

Slup
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These facts may be helpful.

  1. $|G|=p×q$, where $p$ and $q$ are primes and $p<q$. If $p$ doesn't divide $q-1$ then $G$ is cyclic.
  2. $G$ is cyclic $\rightarrow G$ is abelian.
Nitin Uniyal
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