Using the fact that if $a$ and $b$ are two distinct elements of order $5$, either $\langle a \rangle \cap \langle b \rangle =\{e\}$ or $\langle a \rangle = \langle b \rangle$, show that $G$ must have an element of order $3$. Do not assume that $G$ is a cyclic group.
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The last question was concerning a cyclic group, I cannot assume now that it is cyclic. – Algebra May 13 '17 at 16:13
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Have you heard of Lagrange's Theorem? – ancient mathematician May 13 '17 at 16:15
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Yes I've heard of Lagrange's theorem but am struggling to apply it to this particular question – Algebra May 13 '17 at 16:16
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@Algebra We already know that every group of order $15$ is cyclic. So you can use now that it is cyclic. – Dietrich Burde May 13 '17 at 16:17
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@Dietrich Burde- ok thanks – Algebra May 13 '17 at 16:19
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If $|G|=15$ the possible orders of the elements are 1,3,5,15.
If there is an element of order $15$ then its fifth power has order 3. So suppose there are no elements of order 15.
If there are no elements of order 3 then we have 1 element of order 1 and 14 elements of order 5. But the cyclic groups generated by the elements of order 5 do not (as you know) overlap except at the identity. So we will have $1+4k$ elements in total, where $k$ is the number of distinct cyclic subgroups of order 5.
15 is not of this form.
etc.
ancient mathematician
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