# intersections of subgroups

• Sep 19th 2011, 10:09 PM
deniselim17
intersections of subgroups
Let $\displaystyle A$ be a group with normal subgroups $\displaystyle M$ and $\displaystyle N$.
Let $\displaystyle a \in A$.
Suppose $M \cap \langle a \rangle =1$ and $N \cap \langle a \rangle= \langle a^{s} \rangle$ for some positive integer $\displaystyle s$.
Can we find $MN \cap \langle a \rangle$???
• Sep 19th 2011, 10:18 PM
Drexel28
Re: intersections of subgroups
Quote:

Originally Posted by deniselim17
Let $\displaystyle A$ be a group with normal subgroups $\displaystyle M$ and $\displaystyle N$.
Let $\displaystyle a \in A$.
Suppose $M \cap \langle a \rangle =1$ and $N \cap \langle a \rangle= \langle a^{s} \rangle$ for some positive integer $\displaystyle s$.
Can we find $MN \cap \langle a \rangle$???

I'll try to help more if this doesn't work, but I have a hunch that this is a question arising from a larger question. If so, can you give it to us--the context will help.
• Sep 20th 2011, 01:21 AM
Swlabr
Re: intersections of subgroups
Quote:

Originally Posted by deniselim17
Let $\displaystyle A$ be a group with normal subgroups $\displaystyle M$ and $\displaystyle N$.
Let $\displaystyle a \in A$.
Suppose $M \cap \langle a \rangle =1$ and $N \cap \langle a \rangle= \langle a^{s} \rangle$ for some positive integer $\displaystyle s$.
Can we find $MN \cap \langle a \rangle$???

I am not entirely sure what you are asking...I mean, given finite time of course you can! On the other hand, there isn't a definitive answer. For example, take,

$\displaystyle G=C_4\times C_4\times C_4$ ($\displaystyle C_4$ denoting the cyclic group of order $\displaystyle 4$), with your elements being of the form $\displaystyle (i, j, k)$ with $\displaystyle 0\leq i, j, k<4$. Now, let $\displaystyle a=(0, 1, 1)$. Clearly, if $\displaystyle M=\langle (2, 0, 0)\rangle$ then $\displaystyle M\cap \langle a\rangle=1$, while if $\displaystyle N=\langle(0, 2, 2), (0, 2, 1)\rangle$ then $\displaystyle N\cap\langle a\rangle=\langle (0, 2, 2)\rangle=\langle a^2\rangle$. Clearly, $\displaystyle MN\cap \langle a\rangle=\langle a^2\rangle$. On the other hand, if we let $\displaystyle M=\langle(2, 0, 0), (0, 1, 0)\rangle$ then $\displaystyle MN$ contains $\displaystyle (0, 1, 1)$ but $\displaystyle M\cap \langle a\rangle=1$ still...so, basically, it depends on $\displaystyle M$ and $\displaystyle N$ and $\displaystyle \langle a\rangle$, and can be very different depending on different choices...