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Here's the proof of this fact from Schaum's book on group theory -

$1 \in Z(G)$ since $1g = g1$ for all $g \in G$. Consequently $Z(G) \neq \emptyset$.

If $g_1,g_2 \in Z(G)$ and $g \in G$, then $g(g_1 g_2^{-1}) =(g g_1) g_2^{-1} = g_1(g g_2^{-1}) = g_1 g_2^{-1}g$ since $g g_2 = g_2 g$ implies $g_2^{-1} g = g g_2^{-1}$. It follows that $Z(G)$ is a subgroup of $G$.

How does it follow that $Z(G)$ is a subgroup of $G$? I am not sure what this proof has done here to show that $Z(G)$ is a subgroup of $G$. It seems to have demonstrated closure..but I'm not sure if that was the goal as there should be brackets on the final $g_1 g_2^{-1}g$ to make it obvious that that was what they were doing, ie write it as $(g_1 g_2^{-1})g$.

Anyway, could someone explain to me exactly what the process is that they used to show that its a subgroup. Have they implicity demonstrated that all of the group axioms hold?

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The first conclusion should rather read "Consequently $Z(G)\ne \{\}$" (or $\ne \emptyset$), I hope the original book has no $\{0\}$ here. –  Hagen von Eitzen Oct 27 '12 at 20:48
    
Yes the original book has the same notation you used but i dont know the commands for not equal and empty set. –  dukenukem Oct 27 '12 at 20:55
    
I was less worried about the C-style not-equal (\ne by theway) than the weird empty set (\emptyset or \{\} by the way). –  Hagen von Eitzen Oct 27 '12 at 21:04
    
Hehe well i know two new commands now cheers. –  dukenukem Oct 27 '12 at 21:08
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2 Answers

up vote 4 down vote accepted

As you noted, with parentheses added, it would have been easier recognized that he has shown

  • $Z(G)\ne \emptyset$
  • $g_1, g_2 \in Z(G)\Rightarrow g_1 g_2^{-1}\in Z(G)$

Indeed, these two items together show that $Z(G)$ is a subgroup. This "subgroup criterion" should be known; all remaining properties of a group follow from the fact that $G$ is a goup and need not be shown.

By the way, he proves and uses an intermedite result, namely $g_2\in Z(G)\Rightarrow g_2^{-1}\in Z(G)$. In fact, another form of subgroup criterion consists of these three parts:

  • $Z(G)\ne \emptyset$
  • $g_1, g_2 \in Z(G)\Rightarrow g_1 g_2\in Z(G)$
  • $g_1\in Z(G)\Rightarrow g_1^{-1}\in Z(G)$

This form could thus have been used anyway (unless this form of the subgroup criterion is not used at all in the book).

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Ah yes I think I have come across this before as the 'one step subgroup test'. –  dukenukem Oct 27 '12 at 21:06
    
So here is how i read what is happening - $g_1 g_2^{-1}$ is composed of $g_1, g_2^{-1} \in Z(G)$ but mightn't necessarily be in $Z(G)$ itself. But as we show that $g(g_1 g_2^{-1})= (g_1 g_2^{-1})g$ we now can say that $g_1 g_2^{-1}$ is in $Z(G)$ and the one step subgroup test is satisfied. –  dukenukem Oct 27 '12 at 21:47
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One can check to see that a sufficient condition for a subset $H\subseteq G$ to be a subgroup is that $1\in H$ and that $\forall g_1,g_2\in H,\:g_1g_2^{-1}\in H$:

It follows simply by the fact that for any $g\in H$, $g^{-1}=1\cdot g^{-1}\in H$ (since $1\in H$)- thus $H$ is closed under inversion, and likewise for any $g_1,g_2\in H$, $g_2^{-1}\in H$ by the previous argument, and thus $$g_1g_2=g_1(g_2^{-1})^{-1}\in H$$ and consequently $H$ is closed under the group operation. All other group axioms are inherited directly from those of $G$.

What the authors did in the segment you quoted was to show that if $g_1\in Z(G)$ and $g_2\in Z(G)$ then $g_1g_2^{-1}\in Z(G)$, and since $1$ is always in $Z(G)$ the fact that $Z(G)$ is a subgroup of $G$ follows immediately

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