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I want to show that H is a group under multiplication where

$H=\{a^x|x \in \mathbb{R}\}$

To show it's a group then I must show associativity, identity, inverse, and closure.

For Associativiy: Let $ x,y, and z \in H$ then $a^x *(a^y*a^z) \leftrightarrow(a^x*a^y)*a^z$. Thus H is associative.

For identity, is when $x=0$, since $a^0=1$ under multiplication.

Inverse of $a^x$, is $a^{-x}$, since $a^x * a^{-x}=e$.

For closure is where i'm getting a little bit stuck

Let $x,y \in H$, such that $x=a^x, y=a^y$,

Then for closure under multiplication would be $a^x*a^y$ but that would be $a^{x+y}$. Can someone help me out?

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    $\begingroup$ The closure of the operation only means that the product of two elements $a^x$ and $a^y$ in $H$ is of the form $a^z$, for some real number $z$. What you have proven is that taking $z=x+y$ is enough. And just to be VERY pedantic: You must say that $a$ is different from $0$. And even better if you assume $a>0$ and avoid working with complex numbers. $\endgroup$ – Dog_69 Apr 11 at 19:22
  • $\begingroup$ As a side, I would recommend that when you actually prove this out you go through the steps for associativity and not just write that one implies the other. That is the point of the proof, tedious as it may be. $\endgroup$ – JacobCheverie Apr 11 at 19:29
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For closure, you must show that two elements in $H$ will multiply to give another element in $H$. Therefore, let $b, c \in H$ such that $b = a^x$ and $c = a^y$ with $x, y \in \mathbb{R}$. Then, $b * c = (a^x)*(a^y) = a^{x + y}$. Let $z = x + y \in \mathbb{R}$. Then, $b*c = a^z$ with $z \in \mathbb{R}$. Thus, for $b, c \in H$ we have $b*c \in H$ so $H$ is closed under multiplication.

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