[SOLVED] Show that any subgroup of <a> is normal in a dihedral group D_n

**hatsoff** · May 1st 2009, 12:07 PM

Let the dihedral group $D_n$ be given by elements $a$ of order $n$ and $b$ of order $2$ , where $ba=a^{-1}b$ . Show that any subgroup of $<a>$ is normal in $D_n$ .

I have to say I'm not doing a good job of interpreting my textbook, here, which leaves me rather lost to prove what it asks. But here's a shot...

If I read the previous chapter correctly, a dihedral group $D_n=\{a^pb^q\}$ , for all integers $p,q$ with $0\leq p<n$ and $q\in\{0,1\}$ . Hopefully this much is correct.

In order for the subgroup $<a>$ to be normal, we need to show that for each element $g\in D_n$ , it is true that $g^{-1}a^rg=a^s$ for any integer $r\in[0,n)$ and some other integer $s\in[0,n)$ . This clearly holds for all $g=a^pb^0=a^p$ , since $(a^p)^{-1}a^ra^p=a^{n-p}a^ra^p=a^r$ .

So now we need to show that $(a^pb)^{-1}a^r(a^pb)=a^s$ . Now, we know that since $o(b)=2$ then $b^{-1}=b$ . That means $(a^pb)^{-1}=ba^{n-p}$ . So now we evaluate $ba^{n-p}a^ra^pb=ba^rb$ .

Note that since $ba=a^{-1}b=a^{n-1}b$ , then $ba^rb=a^{n-1}ba^{r-1}b$ . And we can get by repeating this:

$ba^rb=a^{n-1}ba^{r-1}b$

$=a^{2n-2}ba^{r-2}b$

$=a^{3n-3}ba^{r-3}b$

$=a^{4n-4}ba^{r-4}b$

...and so on, up to:

$=a^{rn-r}ba^{r-r}b=a^{rn-r}\in D_n$

Q.E.D. ... ?

Have I done this correctly? If not, how did I err? If so, is there a quicker/easier way to do it?

Thanks for your help, guys.

**ThePerfectHacker** · May 1st 2009, 01:11 PM

Originally Posted by hatsoff

I have to say I'm not doing a good job of interpreting my textbook, here, which leaves me rather lost to prove what it asks. But here's a shot...

In general if $G$ is a finite group and $H$ is a subgrop with $(G:H) = 2$ then $H$ is a normal subgroup.
Try proving this instead.

**NonCommAlg** · May 1st 2009, 02:18 PM

the problem is asking you to prove that every subgroup of <a> is normal. let $<a^k>=H$ be any subgroup of $<a>$ and $g \in D_n.$ let $h=a^{kr} \in H.$ we have $ghg^{-1}=(gag^{-1})^{kr}.$

let $g=a^ib^j.$ if $j=0,$ then $gag^{-1}=a$ and thus $ghg^{-1}=a^{kr}=h \in H.$ if $j=1,$ then $g=a^ib=ba^{-i},$ because $ab=ba^{-1}.$ therefore $gag^{-1}=a^{-1}$ and hence $ghg^{-1}=h^{-1} \in H.$

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