0
$\begingroup$

Suppose $G$ and $H$ are finite groups of relatively prime orders. Show that the subgroups $G\times\{e_H\}$ and $\{e_G\}\times H$ of $G\times H$ are characteristic.

My attempt: We want to show that for any automorphism $\phi$ that $\phi(G\times\{e_H\}) = G\times\{e_H\}$ and $\phi(\{e_G\}\times H) = \{e_G\}\times H$. I'm assuming this will rely on the fact that $G$ and $H$ have relatively prime orders, and the order of $G\times H$ is $|G||H|$, but I'm not positive how to prove this.

Any help appreciated!

  • 2
    I suppose you mean $\{e_G\}\times H$ – 2017-02-11
  • 2
    $\newcommand{\Set}[1]{\left\{ #1 \right\}}$Note that you should write $\Set{e_{G}} \times H$. – 2017-02-11

4 Answers 4

2

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$Suppose $\Size{G} = m$ and $\Size{H} = n$.

We show that $G_{1} = G \times \Set{e_{h}} \trianglelefteq G \times H$ is the unique subgroup of order $m$ of $G \times H$. As such, it will be characteristic. A similar argument holds for $H$.

Let $T$ be another subgroup of order $m$. Then $$ \Size{\frac{G_{1} T}{G_{1}}} = \frac{\Size{T}}{\Size{G_{1} \cap T}} \mid \Size{T} = m, $$ but since $G_{1}T/T$ is a subgroup of $G/G_{1} \cong H$, we have also $$ \Size{\frac{G_{1} T}{G_{1}}} \mid \Size{H} = n. $$ Since $\gcd(m, n) = 1$, we have $$ \Size{\frac{G_{1} T}{G_{1}}} = 1, $$ that is, $G_{1} T = G_{1}$, or $T \le G_{1}$, so that $G_{1} = T$.

A slightly more general result is the following

Let $L$ be a finite group, and $N$ a normal subgroup such that $\gcd(\Size{N}, \Size{L/N} ) = 1$. If the subgroup $T$ of $L$ has order dividing the order of $N$, then $T \le N$. In particular, $N$ is unique of its order, and thus characteristic in $L$.

1

Note that the order of $(a,b)\in G\times H$ is the lcm of the orders of $a$ and $b$. We conclude (using the fact that the orders of $G$ and $H$ are coprime primes) that only four orders are possibly

  • $1$ if $a=e_G$, $b=e_H$
  • $|G|$ if $a\ne e_G$, $b=e_H$
  • $|H|$ if $a= e_G$, $b\ne e_H$
  • $|G||H|$ otherwise

Consequently, $G\times\{e_H\}$ is the only non-trivial proper subgroup with an element of order $|G|$; this makes $G\times\{e_H\}$ characterstic, and similar for $\{e_G\}\times H$.

  • 0
    Wait, aren't assuming $G$ and $H$ to be of prime orders here? The assumption is just that they have coprime (= relatively prime) orders. – 2017-02-11
1

A short and simple solution might be the following to prove that for example $G \times \{e_H\}$ is a characteristic subgroup of $G\times H$:

If $(g,1) \in G\times \{e_H\}$ with $g\not=e_G$ has order $m$ then $m$ divides $|G|$ and $\phi((g,1)) = (k,h)$ has order $m$ as well. If $h \not= e_H$ then $h^m \not= e_H$ since otherwise $\text{gcd}(m,|H|) \not= 1$. We conclude that $\phi(G\times \{e_H\}) \subset G\times \{e_H\} $.

0

If $\phi$ is an automorphism of $G \times H$, then $$|\mathrm{Im} \phi \mid_{G \times\{e\}}|$$ divides the order of $G \times H$. Yet there are only two such groups, $G \times \{e\}$ and $H \times \{e\}$ (which can be seen by the other answer, for example.) Yet, because $\phi$ is injective, there is really only one choice.