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I think I'm a bit confused about the order of elements in cyclic groups.

If we suppose $G$ is a group of order 35, and let $x∈G$ such that x≠e, from Lagrange's Theorem $x$ will be of order 5, 7, or 35. If $x$ is of order 35, then $G$ is cyclic and thus has elements of order 5 and 7.

I don't understand the last part where it says that if $x$ is of order 35, $G$ has elements of order 5 and 7. I'd thought that if $G$ is cyclic then all elements of $G$ would have order 35. Obviously this is wrong, but I don't quite understand why?

Any help would be greatly appreciated!

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If every finite group the identity has order $1$. So "all elements of $G$ would have order $35$" can be dismissed as impossible, without knowing anything about $G$. –  Marc van Leeuwen Apr 5 '13 at 18:42

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The order of an elements $g$ in a group $G$ is the smallest number of times that you need to apply the group operation to $g$ to obtain the identity.

Let $G$ be cyclic of order $35$. That means that there is an element $g\in G$ with $g^{35}=e$, and that $g^k\ne e$ for all $1<k<35$. Now, consider $h=g^5$. Then $h^7=(g^5)^7=g^{35}=e$, but $h^k\ne e$ for all $1<k<7$, thus $h$ has order $7$. Similarly, the element $g^7$ has order $5$.

Remark: Cauchy's theorem (which perhaps you did not see yet) states that if $p$ is a prime dividing $|G|$, then $G$ has an element of order $p$. Thus, the only finite groups where all elements except the identity have the same order are $p$-groups, namely groups whose order is a power of a fixed prime $p$. A group of size $35$ is not a $p$-group.

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"the only finite groups where all elements except the identity have the same order are groups of prime order" is not true. There are many Abelian groups where all non-identity elements have order $2$ (and only one of those groups is cyclic); and for other primes there are non-Abelian such groups as well. –  Marc van Leeuwen Apr 5 '13 at 18:46
    
thanks @MarcvanLeeuwen, I meant to write $p$-groups but went on the wrong auto-pilot. Anyways, corrected! –  Ittay Weiss Apr 5 '13 at 18:57
    
Though of course not every p-group has this property (the precise characterization is elementary abelian). –  anon Apr 5 '13 at 21:28

If $G$ is commutative, the order of an element $x$ is the smallest $n \in \mathbb{N}^*$ such that $nx=0_G$.

If $G= \mathbb{Z} / 35 \mathbb{Z}$, $x=\overline{7}$ has order $5$. Indeed,

$2x=\overline{14}$

$3x=\overline{21}$

$4x=\overline{28}$

$5x=\overline{35}=\overline{0}$

$y=\overline{5}$ has order $7$.

We have $\mathbb{Z}/35\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z}$.

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