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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.

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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
@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

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.

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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 $f$.

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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 $\mathbb{Z}/3$.

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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.

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"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
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

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