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Math Help - Prove the LCM of an element in a group

  1. #1
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    Prove the LCM of an element in a group

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

    Attempt at the solution:

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

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

    Attempt at the solution:

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

    Let (a_1, a_2, . . . , a_n) \in G_1 \times G_2 \times ... \times G_n. Then |(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 \left(g_1,\cdots,g_n\right)^x=\left(g_1^x,\cdots,g  _n ^x\right) and so clearly \left(g_1,\cdots,g_n\right)^{\text{lcm}\left(|a_1|  ,\cdots,|a_n|\right)}=(e,\cdots,e) and so \left|\left(|g_1|,\cdots,|g_m|\right)\right|\mid \text{lcm}\left(|g_1|,\cdots,|g_n|\right) by basic group theory. But, 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 1\leqslant k\leqslant n and so by definition \text{lcm}\left(|g_1|,\cdots,|g_n|\right)\mid \left|\left(g_1,\cdots,g_n\right)\right| and so it follows that \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 \pm is positive and the order of something must be positive it is clear that we must have \left|\left(g_1,\cdots,g_n\right)\right|=\text{lcm  }\left(|g_1|,\cdots,|g_n|\right)
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