# Thread: Prove the LCM of an element in a group

1. ## 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?

2. Originally Posted by crushingyen
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 $\left|\left(g_1,\cdots,g_n\right)\right|=\text{lcm }\left(|g_1|,\cdots,|g_n|\right)$