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I am working on some homework for modern algebra class. The problem I just finished seems relatively easy, but I have learned to be wary of that feeling when it comes to this material. Below are the problem statement and my solution. Please comment on my solution's validity, if you are so inclined.

This is problem 3, pg. 92 of Herstein's Abstract Algebra. Let $G$ be a group, and $A = G \times G = \{ (x,y) |\ \ x,y \in G \} $ the direct product of that group with itself. Let $T = \{(g,g) |\ \ g \in G\}$. What I need to do is to prove that

  1. $T \cong G$ and
  2. $T \triangleleft A \Leftrightarrow G\ $ is abelian.

My solutions for these are as follows:

  1. Let $\varphi : T \rightarrow G |\ \ \varphi((g,g)) = g$. We can show that $\varphi$ is an isomporphism in the following way: first, we will prove that $\varphi$ is a homomorphism.
    Let $t,s \in T |\ \ t = (g_{1},g_{1}), s = (g_{2},g_{2}), g_{1}, g_{2} \in G$.
    Then $\varphi(st) = \varphi((g_{1}g_{2},g_{1}g_{2})) = g_{1}g_{2} = \varphi((g_{1},g_{1}))\varphi((g_{2},g_{2})) = \varphi(s)\varphi(t)$, indicating that $\varphi$ is a homomorphism.
    Next we will show that $\varphi$ is a monomorphism. Let $\varphi(s) = \varphi(t)$. This implies that $\varphi((g_{1},g_{1}))=\varphi((g_{2},g_{2})) \Rightarrow g_{1} = g_{2}$, proving that $\varphi$ is a 1-1 mapping, and so is a monomorphism.
    Finally, we must show that $\varphi$ is onto. We can show that this mapping covers all of $G$ like so: $\forall\ g \in G\ \exists\ x \in T |\ \ x = (g,g)\ \&\ \varphi(x) = g$. This proves that $\varphi$ is an isomorphism from $T$ to $G$. The existence of this isomorphism proves that $T \cong G$. $\blacksquare$
  2. First, we will prove that $T \triangleleft A \Rightarrow G$ is abelian. $T \triangleleft A \Rightarrow \forall\ x \in A, x^{-1}Tx \subset T$. Let $x = (g_{1},g_{2}) \in A, g_{1},g_{2} \in G, t = (h,h) \in T, h \in G$. Then $x^{-1} = (g_{1}^{-1},g_{2}^{-1})$, with both inverses existing as a consequence of $A$ being a group. This is easy to see since $A$ is the direct product of the group $G$ with itself. Let us now form the product: $x^{-1}tx = (g_{1}^{-1}hg_{1}, g_{2}^{-1}hg_{2}) = u \in T$, since $T \triangleleft A$. This, in turn, tells us about the struture of the product; because of $T$'s definition, we can tell that $g_{1}^{-1}hg_{1} = g_{2}^{-1}hg_{2}$. Rearranging the terms, this gives us $h = g_{1}g_{2}^{-1}hg_{2}g_{1}^{-1}.$ We know that $(g_{1}g_{2}^{-1})^{-1} = g_{2}g_{1}^{-1}$, so we can multiply this through: $g_{2}g_{1}^{-1}h = hg_{2}g_{1}^{-1}$, which shows that $G$ must be abelian.
    Next, we will show that the other implication is true and prove that $G$ being abelian implies that $T \triangleleft A$. Let $x=(g_{1},g_{2}) \in A, t = (h,h) \in T, g_{1},g_{2},h \in G$. This makes the product $x^{-1}hx = (g_{1}^{-1}hg_{1},g_{2}^{-1}hg_{2})$. We know that $G$ is abelian, so $(g_{1}^{-1}hg_{1},g_{2}^{-1}hg_{2}) = (g_{1}^{-1}g_{1}h,g_{2}^{-1}g_{2}h) = (h,h) \in T$, proving that $x^{-1}Tx \subset T\ \forall\ x \in A$, and, equivalently, that $T \triangleleft A$.
    These two statements together prove that $T \triangleleft A \Leftrightarrow G\ $ is abelian. $\blacksquare$.

Thanks in advance for any help you may be able to provide.

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2 Answers 2

up vote 1 down vote accepted

This looks pretty good. I only have two comments:

  1. Your $\varphi$ could also be introduced as the restriction of one of the projection maps that come with $G \times G$. Depending on what you have already proved about products of groups, this might save you some work. Also note that $\varphi$ has an inverse that is easy to write down: define $\Delta\colon G \to T$ by $\Delta(g) = (g, g)$.

  2. When proving the "$\Rightarrow$" direction in the second part, you should conclude by saying something like, "set $g_1 = e$ to obtain $g_2h = hg_2$ for all $h, g_2 \in G$". Overall, it might be quicker to write, for each $g, h \in G$, $$ (h, e)(g, g)(h, e)^{-1} = (h, e)(g, g)(h^{-1}, e) = (hgh^{-1}, g) \in T. $$ Then manipulate $hgh^{-1} = g$ to finish the proof.

But in a technical sense I think you've improved a great deal.

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Thanks to both you and @AMPerrine for your helpful replies. I'll clean it up a bit soon with the feedback you've provided. –  karmic_mishap Oct 30 '11 at 22:22

First off, I believe your proofs to be correct. However, I suggest making the first part of the second proof a bit cleaner. Try this:

Let $x,y$ be elements of $G$, and $e$ the identity. Then $(e,x)\in A$ and $(y,y)\in T$.

Proceed with your proof as above and you should get your result with much less fuss.

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