# Let $G$ be a group of order 35. Show that $G \cong Z_{35}$ [duplicate]

Let $G$ be a group of order 35. Show that $G \cong Z_{35}$

Now, I am going to assume that this is a LaGrange based question. Also, I know that in order to be isomorphic, it must be one to one and onto.

• Do you know the Sylow theorems? Mar 21, 2015 at 19:54
• @MattSamuel, Im just learning them. I havnt really worked problems using them yet
– cele
Mar 21, 2015 at 19:58
• Try looking here. I believe this is a duplicate.
– Eoin
Mar 21, 2015 at 19:59
• @Cele Be careful about pronouns. When you say "it must be one to one and onto" you should be very clear about what "it" refers to (a homomorphism) Mar 21, 2015 at 20:05

$\#G = 5.7$ Let ${S}_{5}$ denote the Sylow 5-subgroup and ${S}_{7}$ the Sylow 7-subgroup. Now $${S}_{5} \cap {S}_{7}$$ is a subgroup and the number of elements in it divides 5 and 7 (Lagrange). So $$\#{S}_{5} \cap {S}_{7} = 1$$ Then $$\#{S}_{5}{S}_{7} = \frac{\#{S}_{5}.\#{S}_{7}}{ \#{S}_{5} \cap {S}_{7}}=5.7=35$$ Because ${S}_{5}{S}_{7} \subset G$ and $\#{S}_{5}{S}_{7} = \#G$ it follows that: $$G = {S}_{5}{S}_{7}$$ By Sylow's second theorem (because 5 and 7 are primes):
${S}_{5}$ is the only Sylow 5-subgroup of $G$. The same for ${S}_{7}$. So $${S}_{5},{S}_{7} \triangleleft G$$ This are all the requirements that needed to be true for a group to be isomorphic with the direct product: ${S}_{5} \times {S}_{7}$. Since ${S}_{5}$ and ${S}_{5}$ are cyclic (number of elements is prime) it follows that ${S}_{5} \cong \mathbb{Z}_{5}$. Same for ${S}_{7}$. So ${S}_{7} \cong \mathbb{Z}_{7}$. Because 5 and 7 have no common divisors other than 1 it follows: $$\mathbb{Z}_{5} \times \mathbb{Z}_{7} \cong \mathbb{Z}_{35}$$ And so $$G \cong {S}_{5} \times {S}_{7} \cong \mathbb{Z}_{35}$$
• Yes the number of elements of G. Sorry I always use $\#$ forgot that it might be unclear. Mar 21, 2015 at 20:16
• I am assuming that right above your last 'paragraph,' that is suppose to be $S_5,S_7 \triangleleft G$?