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Let $G$ and $H$ be groups and $G\times H$ their direct product.

a) Prove that $\{(x,e) : x\in G\}$ is a subgroup of $G\times H$

b) Prove that $\{(x,x) : x\in G\}$ is a subgroup of $G\times G$

I don't know where to begin. Do you still have to prove the resulting subgroup is closed under the operation or just closed under inverses.

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    $\begingroup$ you have to prove both.... $\endgroup$
    – user87543
    Commented Oct 14, 2013 at 5:04

1 Answer 1

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One easy way to prove a subset $K \subset M$ is a subgroup of a group $M$ is the Subgroup test : If $a,b \in K$, then show that $ab^{-1} \in K$.

Consider $K := \{(x,e) : x\in G\} \subset G\times H =: M$. Then, for $a = (x_1, e), b = (x_2,e) \in K$, note that $$ ab^{-1} = (x_1x_2^{-1},e) \qquad\text{ (why?)} $$ Hence, $ab^{-1} \in K$, and so $K < M$.

Can you try the same thing for $K:= \{(x,x) : x\in G\} \subset G\times G =: M$

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  • $\begingroup$ If i am using the two step subgroup test can I use similar steps and say ab is (x1x2,e)∈K and a^-1 is (x^-1,e)∈K therefore it is a subgroup? For the other part to the question, K:={(x,x):x∈G}⊂G×G=:M my instinct would be to let a=(x1,x2) and b=(x3,x4) then ab would then be (x1x3,x2x4)∈K and a^-1 is (x1^-1,x2^-1)∈K. Am I using proper reasoning? $\endgroup$
    – Student
    Commented Oct 14, 2013 at 17:40
  • $\begingroup$ Well, you need $x_1 = x_2$ and $x_3=x_4$ for this to make sense, right? $\endgroup$ Commented Oct 14, 2013 at 18:05
  • $\begingroup$ If i make that distinction known then it would be correct? With that then x1x3=x2x4 in ab which would make sense and same for a^-1. $\endgroup$
    – Student
    Commented Oct 14, 2013 at 18:11

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