Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It seems the definition of the center of a group and a normal subgroup are the same so I'm wondering what the difference is between the two?

A group $H$ is normal in $G$ iff $Hg=gH$ for all $g \in G$.

The center of a group $Z(G) = \{z| \in G$ and for all $g \in G, gz=zg\}$

Those statements seem equivalent to me.

share|improve this question
    
You should edit your question to give the definitions as you understand them. –  Chris Eagle Oct 27 '12 at 19:17
    
The definitions are actually rather different; please add the definitions that you’re using, so that we can try to pin down just what it is that you’re missing. –  Brian M. Scott Oct 27 '12 at 19:20
    
Ok I've edited in the definitions I have. –  James Oct 27 '12 at 19:22
2  
@James: Even though $gH = Hg$, this does NOT imply $gh = hg$ for all $h \in H$. What $gH = Hg$ does imply is that for any element $h \in H$, $gh = h_0g$ for some $h_0 \in H$. See the difference? –  Mikko Korhonen Oct 27 '12 at 19:25
    
@m.k. Oh..I think I get it..have I got this right - With Z(G), gz = zg a specific z is needed to satisfy the equality..whereas with a normal subgroup, gH = Hg means any h in H can satisfy the equality? –  James Oct 27 '12 at 19:31

4 Answers 4

The difference is that $Hg = gH$ means that $ \forall h \in H, \forall g\in G, gh \in Hg$ and $ hg \in gH$. Note that it does not require that $gh = hg$, just that it is in the right coset.

On the other hand, for an element $h \in Z(G), \forall g \in G, hg = gh$ This is a stronger condition. As such the centre is always a normal subgroup, but not all elements of normal subgroups are in the centre.

share|improve this answer
    
"means that.." what you're saying there is $gH \subseteq Hg$, but there are examples of groups with $gH \subseteq Hg$ but $gH \neq Hg$. –  Mikko Korhonen Oct 27 '12 at 19:55
    
Like what? It wouldn't work for finite $H$ since $gh_1 = h_2g, \forall h_1 \in H$ defines an injective map from $gH \to Hg$ and then this is a bijection since H is finite. I'm sure you're right, just would like to see how it goes wrong! –  Tom Oldfield Oct 27 '12 at 20:51
2  
Yes, you're right, a counterexample $H$ would need to be infinite. See this question where you can find examples of $H$ such that $aHa^{-1} \subset H$ yet $aHa^{-1} \neq H$, equivalently examples of $aH \subset Ha$ yet $aH \neq Ha$. –  Mikko Korhonen Oct 27 '12 at 20:56

If $x\in Z(G)$, you have that $g^{-1}xg = x$ for every $g\in G$, whereas if $H$ is normal and $x \in H$, you only have that $g^{-1}xg \in H$. This is a much weaker condition.

In other words, the center is invariant pointwise under conjugation by $G$, whereas in general normal subgroups are only invariant under conjugation as a whole subgroup.

share|improve this answer

The statements are not equivalent. What you’re missing is that $Hg=gH$ does not imply that $hg=gh$ for all $h\in H$: the set of elements $\{hg:h\in H\}$ can be equal to the set of elements $\{gh:h\in H\}$ without each of the individual products $hg$ and $gh$ being the same.

For a concrete example of this, let $G=S_3$, the symmetry group of an equilateral triangle; you can see its multiplication table here. Let $H=\{e,d,f\}$; it’s easy to check that $H$ is a subgroup of $G$. Then $aH=\{a,b,c\}=Ha$, but $ad=b\ne c=da$. You can go on to check that $xH=Hx$ for every $x\in G$, so that $H$ is normal in $G$, but none of the elements $a,b$, and $c$ commutes with $d$ or $f$.

share|improve this answer

The center is a normal subgroup, but there are normal subgroups which are different from the center.

For example consider a cyclic group $\mathbb{Z} /6$, since $\mathbb{Z}/6$ is abelian the definition of the center you gave tells us that $Z(\mathbb{Z}/6) = \mathbb{Z}/6$. However there are also normal subgroups $\mathbb{Z}/2$ and $\mathbb{Z}/3$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.