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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.

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    $\begingroup$ Do you know the Sylow theorems? $\endgroup$ Mar 21, 2015 at 19:54
  • $\begingroup$ @MattSamuel, Im just learning them. I havnt really worked problems using them yet $\endgroup$
    – cele
    Mar 21, 2015 at 19:58
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    $\begingroup$ Try looking here. I believe this is a duplicate. $\endgroup$
    – Eoin
    Mar 21, 2015 at 19:59
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    $\begingroup$ @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) $\endgroup$ Mar 21, 2015 at 20:05

1 Answer 1

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$\#G = 5.7$ 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 $$G \cong {S}_{5} \times {S}_{7} \cong \mathbb{Z}_{35}$$

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    $\begingroup$ in regard to your notation, I am assuming that # refers to the order?? $\endgroup$
    – cele
    Mar 21, 2015 at 20:14
  • $\begingroup$ Yes the number of elements of G. Sorry I always use $\#$ forgot that it might be unclear. $\endgroup$
    – abcdef
    Mar 21, 2015 at 20:16
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    $\begingroup$ I am assuming that right above your last 'paragraph,' that is suppose to be $S_5,S_7 \triangleleft G$? $\endgroup$
    – cele
    Mar 21, 2015 at 21:01
  • $\begingroup$ Yeah indeed, my mistake $\endgroup$
    – abcdef
    Mar 21, 2015 at 21:29
  • $\begingroup$ Hi what theorem are you using when you say that these "are all the requirements needed for a group to be isomorphic with the direct product"? $\endgroup$
    – Alex.F
    May 11, 2018 at 23:12

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