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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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    $H$ is normal if whenever you take a thing in $H$ and conjugate it with anything in $G$, you still get a thing in $H$, but maybe a different thing in $H$ than what you started with. $Z$ is the center if whenever you take a thing in $Z$ and conjugate with anything in $G$, you get the same exact thing back.2012-11-20

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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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    Yes, you're right, a counterexample $H$ would need to be infinite. See [this question](http://math.stackexchange.com/questions/107862/conjugate-subgroup-strictly-contained-in-the-initial-subgroup/107874#107874) 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 $aH \neq Ha$.2012-10-27
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If N is a normal subgroup of G $ N◁G $ , then $ gng‾ ∈N ;∀g∈G $ or simply $ gNg‾ = N $ It follows that for a Normal subgroup it's left and right cosets are same . $ gN = Ng ; \ for \ ∀g∈G $ It does not however imply for some $ n_1 ∈N $ that $ gn_1 = n_1g $ . It rather implies that there is some $ n_2∈N $ such that $ gn_1 = n_2g $ Here the Group is unaffected by conjugation.

Center of Group $Z(G)$ is defined as set of all elements $ \{s: gs=sg ; ∀g∈G\} $ And that set is a group in itself thus is a subgroup. For center of Group $Z(G)$ and $s∈Z(G)$ it follows that $ gsg– =s $ Here each element in itself is unaffected by conjugation.