# intersections of subgroups

Printable View

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

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