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Thread: Prove the LCM of an element in a group

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    Prove the LCM of an element in a group

    If $\displaystyle G_1, G_2, . . . , G_n$ are finite groups, prove that the order of $\displaystyle (a_1, a_2, . . . , a_n)$ in $\displaystyle G_1 \times G_2 \times ... \times G_n$ is the LCM of the orders $\displaystyle |a_1|, |a_2|, ..., |a_n|$.

    Attempt at the solution:

    Since $\displaystyle G_1, G_2, . . . , G_n$ are finite groups, we know by definition that $\displaystyle G_1 \times G_2 \times ... \times G_n$ has order $\displaystyle |G_1| \cdot |G_2| \cdot \cdot \cdot |G_n|$.

    Let $\displaystyle (a_1, a_2, . . . , a_n) \in G_1 \times G_2 \times ... \times G_n$. Then $\displaystyle |(a_1, a_2, . . . , a_n)| = |a_1| \cdot |a_2| \cdot \cdot \cdot |a_n|$.

    From this point I think it's obvious enough that it's the LCM, but I don't know if it's sufficient. Could anyone help me finish this to make it more formal if possible?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by crushingyen View Post
    If $\displaystyle G_1, G_2, . . . , G_n$ are finite groups, prove that the order of $\displaystyle (a_1, a_2, . . . , a_n)$ in $\displaystyle G_1 \times G_2 \times ... \times G_n$ is the LCM of the orders $\displaystyle |a_1|, |a_2|, ..., |a_n|$.

    Attempt at the solution:

    Since $\displaystyle G_1, G_2, . . . , G_n$ are finite groups, we know by definition that $\displaystyle G_1 \times G_2 \times ... \times G_n$ has order $\displaystyle |G_1| \cdot |G_2| \cdot \cdot \cdot |G_n|$.

    Let $\displaystyle (a_1, a_2, . . . , a_n) \in G_1 \times G_2 \times ... \times G_n$. Then $\displaystyle |(a_1, a_2, . . . , a_n)| = |a_1| \cdot |a_2| \cdot \cdot \cdot |a_n|$.

    From this point I think it's obvious enough that it's the LCM, but I don't know if it's sufficient. Could anyone help me finish this to make it more formal if possible?
    First note of course that $\displaystyle \left(g_1,\cdots,g_n\right)^x=\left(g_1^x,\cdots,g _n ^x\right)$ and so clearly $\displaystyle \left(g_1,\cdots,g_n\right)^{\text{lcm}\left(|a_1| ,\cdots,|a_n|\right)}=(e,\cdots,e)$ and so $\displaystyle \left|\left(|g_1|,\cdots,|g_m|\right)\right|\mid \text{lcm}\left(|g_1|,\cdots,|g_n|\right)$ by basic group theory. But, $\displaystyle g_k^{\left|\left(g_1,\cdots,g_n\right)\right|}=e\i mplies |g_k|\mid\left|\left(g_1,\cdots,g_n\right)\right|$ for ever $\displaystyle 1\leqslant k\leqslant n$ and so by definition $\displaystyle \text{lcm}\left(|g_1|,\cdots,|g_n|\right)\mid \left|\left(g_1,\cdots,g_n\right)\right|$ and so it follows that $\displaystyle \left|\left(g_1,\cdots,g_n\right)\right|=\pm\text{ lcm}\left(|g_1|,\cdots,|g_n|\right)$ and since the RHS sans the $\displaystyle \pm$ is positive and the order of something must be positive it is clear that we must have $\displaystyle \left|\left(g_1,\cdots,g_n\right)\right|=\text{lcm }\left(|g_1|,\cdots,|g_n|\right)$
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