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Let $H$ and $K$ be subgroups of a group $G$ and assume that $\gcd(|H|,|K|)=1$. Show that $H \cap K={1_G}$.

Approach: I don't know. I feel like we have to use lagrange's theorem and something about the order of an element $1_G\neq a\in H \cap K$ to generate a contradiction.

Let see if $\exists a \in H \cap K$, then $a\in H$ and $a\in K$ We also know automatically that $<a>$ is a subgroup of $H$ and a subgroup of $K$, but by lagrange's theorem $ord(a)$ divides $|H|$ and $|K|$ which can't happen because $|H|$ and $|K|$ are relatively prime. Something like that.

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$H\cap K$ is subgroup of $H$ and $K$. Thus, by Lagrange's theorem, $|H\cap K|$ divides $|H|$ and $|K|$, but $gcd(|H|,|K|)=1$ and therefore $|H\cap K|=1$, i.e., $H\cap K=1_{G}$.

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  • $\begingroup$ Thanks, can you help me with the following one ? math.stackexchange.com/questions/1890778/… $\endgroup$ – TheMathNoob Aug 13 '16 at 7:33
  • $\begingroup$ ....heyy, almost six months have passed and I have no memories of group theory. I feel like I am reading Chinese when I read this post. XDDDDDDDDDDDDD. I feel like my own attempt is Chinese to me. $\endgroup$ – TheMathNoob Mar 19 '17 at 1:31
  • $\begingroup$ what is a group lol? what is math? $\endgroup$ – TheMathNoob Mar 19 '17 at 1:37

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