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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.

cele
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1 Answers1

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

abcdef
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    in regard to your notation, I am assuming that # refers to the order?? – cele Mar 21 '15 at 20:14
  • Yes the number of elements of G. Sorry I always use $#$ forgot that it might be unclear. – abcdef Mar 21 '15 at 20:16
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    I am assuming that right above your last 'paragraph,' that is suppose to be $S_5,S_7 \triangleleft G$? – cele Mar 21 '15 at 21:01
  • Yeah indeed, my mistake – abcdef Mar 21 '15 at 21:29
  • 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"? – Alex.F May 11 '18 at 23:12
  • I don't know the name, but the isomorphism is given by $\sigma: S_5 \times S_7 \rightarrow G: (g,h) \mapsto gh$. First show commutativity: $\sigma(g,h)=gh =hg$. To see this, let $x \in G$ such that $gh = hg x$. Then $x = g^{-1}h^{-1}gh \in S_5 \cap S_7$, which can be seen since both subgroups are normal. But then $x = e$. The same argument shows that $\sigma$ is injective. Now use that $|S_5 \times S_7| = 35 = |G|$ to see that $\sigma$ is also surjective. – abcdef May 12 '18 at 07:52