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Let $a,b$ be elements in a group $G$ such that $|a|=25$ and $|b|=49$. Prove that $G$ contains an element of order 35.

If $G$ is finite then we can say that $5,7 \mid G$ and hence by Cauchy's Theorem, we have elements $c,d$ in $G$ of orders 5 and 7, respectively. So we can take $cd$ to be the element of order 35. What about the case when $G$ is infinite?

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    There are finite counterexamples as well. The companion matrix of the 49th cyclotomic polynomial (of degree 42) over Z/25Z has no eigenvalues that are 7th roots of unity, and in particular, its 7th power acts fixed point freely on (Z/25Z)^42, and the semi-direct product has no elements of order 35, but elements of orders 25 and 49. I think reversing the roles of$5$and$7$gives a smaller group, but the argument is a little more complicated as the 25th cyclotomic polynomial is reducible mod 49.2012-04-04

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As stated, the claim is false: let $G=C_{25}*C_{49}$, the free product of a cyclic group of order $25$ and a cyclic group of order $49$; if $a$ is a generator of $C_{25}$ and $b$ is a generator of $C_{49}$, then $a$ and $b$ have the desired properties.

However, in a free product $H*K$, the only elements of finite order are conjugate to elements of finite order in $H$ or in $K$; so in $C_{25}*C_{49}$, the only elements of finite order are conjugate to an element of $C_{25}$ or to an element of $C_{49}$, and thus must have order $1$, $5$, $25$, $7$, or $49$; no element of $G$ has order $35$.

So presumably, there are some extra hypothesis on $G$ to make the implication true...

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    @Gerry, or G finite and nilpotent ...2012-04-04