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If the index and order of a normal subgroup and subgroup are relatively prime, then the subgroup is contained in the normal subgroup

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.18

Let $G$ be a group and let $H,N \leq G$ with $N$ normal in $G$. Prove that if $|H|$ and $[G:N]$ are relatively prime then $H \leq N$.

Solution: First we prove a lemma.

Lemma: Let $G$ be a group, $H \leq G$, and $x \in G$ an element of finite order $n$. If $k$ is the least positive integer such that $x^k \in H$, then $k|n$.

Proof: If $k$ does not divide $n$, we have $n = qk+r$ for some $0 < r < k$ by the division algorithm. Now $1 = x^n = x^{qk}x^r \in H$, and $x^{qk} = (x^k)^q \in H$. Thus $x^r \in H$, which contradicts the minimality of $k$. Thus $k|n$. $\square$

Now to the main result.

Suppose $x \in H$, and let $k$ be the least positive integer such that $x^k \in N$. ($k$ exists since $H$ is finite.) By a previous exercise, as an element of $G/N$, $|xN| = k$, so that $k$ divides $[G:N]$. Moreover, we have $|x|$ divides $|H|$ by Lagrange's Theorem, so that (by the lemma) $k$ divides $|x|$ and thus divides $|H|$. Because $|H|$ and $[G:N]$ are relatively prime, then, $k = 1$. But then $|xN| = 1$, so $xN = N$, and we have $x \in N$. So $H \subseteq N$. By a previous exercise $H \leq N$.


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